# NCERT Solutions Class 12 Mathematics Chapter 7

NCERT Solutions for Class 12 Mathematics Chapter 7 Integrals introduces 'Integration,' the other side of differentiation. Differentiation provides us with the rate of change, or in geometric terms, the slope of the tangent, at every given point in the curve, while Integration gives us the area under the curve. Integrals are the inverse of differentiation; therefore, we can get the original function back by integrating the resultant derived from differentiating a function. NCERT solutions Class 12 Mathematics Chapter 7 explains what Integration is and how to use it, as well as the numerous methods for calculating it.

The integration method allows you to sum infinitesimally small parts infinitely many times, allowing you to calculate the area under the curve. With the help of examples and easily comprehensible methods provided by Extramarks, students find the concept intuitive and are able to compute integration quickly. Chapter 7 Mathematics Class 12 Integrals teaches students about integration, Its types (indefinite and definite) as well as their relationship to differentiation. It also outlines multiple approaches that can be used to achieve integration. Integration is a useful method for resolving scientific and technical issues. It can also be used to answer questions in economics, finance, and probability.

### Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 7

Integration, a process opposite of differentiation, consists of functions known as integrals that must satisfy a given differential equation. Students shall be responsible for identifying the functions whose differential will be given to them. Here are some concepts that will be covered in this paper.

• Integration
•  Indefinite integrals
• Application of integrals
• Integrals for class 12
• Application of integrals for class 12
• Integral calculus

Indefinite integrals, geometrical interpretation of indefinite integrals, properties of indefinite integrals, standard integrals, and methods of integration including integration by substitution method, integration by partial fractions, and integration by parts are among the significant topics covered in NCERT Solutions Class 12 Mathematics Chapter 7. Definite integrals, the fundamental theorem of calculus, and the properties of definite integrals are some of the other essential topics covered in this chapter as well. The goal of curating these solutions is to promote fundamental knowledge of integral calculus to assist students in their studies.

### List of NCERT Solutions Class 12 Mathematics Chapter 7 Exercises & Answer Solutions

One of the most important topics in calculus - integration, has a wide range of practical applications. This lesson contains several formulae and hence necessitates laser-sharp focus. Revising the solutions regularly is the greatest approach to remember these concepts and measure your understanding of integration. The following is an exercise-by-exercise detailed analysis of NCERT Solutions Class 12 Mathematics Chapter 7 Integrals to assist students in developing a strong understanding of this subject:

Chapter 7 Ex 7.1 - 22 Questions - Class 12 Mathematics

Chapter 7 Ex 7.2 - 39 Questions - Class 12 Mathematics

Chapter 7 Ex 7.3 - 24 Questions - Class 12 Mathematics

Chapter 7 Ex 7.4 - 25 Questions - Class 12 Mathematics

Chapter 7 Ex 7.5 - 23 Questions - Class 12 Mathematics

Chapter 7 Ex 7.6 - 24 Questions - Class 12 Mathematics

Chapter 7 Ex 7.7 - 11 Questions - Class 12 Mathematics

Chapter 7 Ex 7.8 - 6 Questions - Class 12 Mathematics

Chapter 7 Ex 7.9 - 22 Questions - Class 12 Mathematics

Chapter 7 Ex 7.10 - 10 Questions - Class 12 Mathematics

Chapter 7 Ex 7.11 - 21 Questions - Class 12 Mathematics

Chapter 7 Miscellaneous Exercise - 44 Questions - Class 12 Mathematics

Our subject experts create NCERT Solutions so as to assist students in understanding concepts more quickly and correctly. NCERT Solutions provide extensive, step-by-step explanations of textbook difficulties. NCERT  Solutions for all primary, secondary and higher secondary classes –

• NCERT Solutions class 1
• NCERT Solutions class 2
• NCERT Solutions class 3
• NCERT Solutions class 4
• NCERT Solutions class 5
• NCERT Solutions class 6
• NCERT Solutions class 7
• NCERT Solutions class 8
• NCERT Solutions class 9
• NCERT Solutions class 10
• NCERT Solutions class 11
• NCERT Solutions class 12

NCERT Solutions Class 12 Mathematics Chapter 7 Formula List

NCERT solutions Class 12 Mathematics chapter 7 includes various important concepts necessary for mastering higher-level mathematics. Students must understand the derivation of the principles and formulas given. This will assist them in implementing the measures necessary to solve integration problems and limit their mistakes. Students should also keep a formula chart to quickly and efficiently review the formulae before exams. The following are some key formulae mentioned in NCERT Solutions for Class 12 Mathematics Chapter 7:

• ∫ f(x) dx = F(x) + C
• Power Rule: ∫ xn dx = (xn+1)/ (n+1)+ C. (Where n ≠ -1)
• Exponential Rules: ∫ ex dx = ex + C
• ∫ ax dx = ax /ln(a) + C
• ∫ ln(x) dx = x ln(x) -x + C
• Constant Multiplication Rule: ∫ a dx = ax + C, where a is the constant.
• Reciprocal Rule: ∫ (1/x) dx = ln(x)+ C
• Sum and Difference Rules:
• ∫ [f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
• ∫ [f(x) - g(x)] dx = ∫f(x) dx - ∫g(x) dx
• ∫ k f(x) dx = k ∫f(x) dx, where k is any real number.
• Integration by parts: ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫[d/dx f(x) * ∫ g(x) dx]dx
• ∫ cos x dx = sin x + C
• ∫ sin x dx = -cos x + C
• ∫ sec 2x dx = tan x + C
• ∫ cosec 2x dx = -cot x + C
• ∫ sec x tan x dx = sec x + C
• ∫ cosec x cot x dx = - cosec x + C

NCERT Class 12 Mathematics Syllabus CBSE

Term - 1

 Unit Name Chapter Name Relations and Function Relations and Functions Inverse Trigonometric Functions Algebra Matrices Determinants Calculus Continuity and Differentiability Application of Derivatives Linear Programming Linear Programming

Term - 2

 Unit Name Chapter Name Calculus Integrals Application of Integrals Differential Equations Vectors and Three-Dimensional Geometry Vector Algebra Three Dimensional Geometry Probability Probability

NCERT Class 12 Mathematics Exam Pattern

 Duration of Marks 3 hours 15 minutes Marks for Internal 20 marks Marks for Theory 80 marks Total Number of Questions 38 Questions Very short answer question 20 Questions Short answer questions 7 Questions Long Answer Questions (4 marks each) 7 Questions Long Answer Questions (6 marks each) 4 Questions

Key features NCERT Mathematics class 12th chapter 7

Students can study the following topics by learning the NCERT Solutions for Integrals:

As the inverse of differentiation, integration Substitution, partial fractions, and parts are used to integrate a number of functions. Simple integrals of the following categories and problems dependent on them are evaluated. Fundamental Theorem of Calculus, Definite Integrals as a Limit of a Sum (without proof). Evaluation of definite integrals and basic properties of definite integrals.

NCERT Exemplar Class 12 Mathematics

All solutions and problems are given to help students prepare for their final exams. These example questions are a little more complex, and they cover each and every concept covered in each chapter of the Class 12 Mathematics subject. Students will fully understand all the concepts covered in each chapter by practising these NCERT exemplar for Mathematics Class 12. Exemplars provide the best solutions to challenges that students confront. To match the ideas taught in each class and provide the greatest practising materials or worksheets for students, all of these questions strictly follow updated 2022-23 CBSE guidelines.

Q.1

$\mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{\mathrm{cosx}-\mathrm{sinx}}{1+\mathrm{sin}2\mathrm{x}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{cosx}-\mathrm{sinx}}{1+\mathrm{sin}2\mathrm{x}}=\frac{\mathrm{cosx}-\mathrm{sinx}}{{\mathrm{sin}}^{2}\mathrm{x}+{\mathrm{cos}}^{2}\mathrm{x}+2\mathrm{sinxcosx}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cosx}-\mathrm{sinx}}{{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^{2}}\\ \therefore \int \frac{\mathrm{cosx}-\mathrm{sinx}}{1+\mathrm{sin}2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \frac{\mathrm{cosx}-\mathrm{sinx}}{{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}=\mathrm{sinx}+\mathrm{cosx}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}-\mathrm{sinx}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{cosx}-\mathrm{sinx}}\\ \int \frac{\mathrm{cosx}-\mathrm{sinx}}{1+\mathrm{sin}2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\mathrm{cosx}-\mathrm{sinx}}{{\left(\mathrm{t}\right)}^{2}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{\mathrm{cosx}-\mathrm{sinx}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{t}}^{-2}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{\mathrm{t}}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{\mathrm{sinx}+\mathrm{cosx}}+\mathrm{C}\end{array}$

.

Q.2 Find the integral of the function tan3 2x sec 2x.

Ans.

$\begin{array}{l}\int {\mathrm{tan}}^{3}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{tan}}^{2}2\mathrm{xtan}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{tan}}^{2}2\mathrm{xtan}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \left({\mathrm{sec}}^{2}2\mathrm{x}-1\right)\mathrm{tan}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ =\int {\mathrm{sec}}^{2}2\mathrm{x}.\mathrm{tan}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \mathrm{tan}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}t}=\mathrm{sec}2\mathrm{x}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{sec}2\mathrm{xtan}2\mathrm{x}\\ ⇒\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{sec}2\mathrm{xtan}2\mathrm{x}}\\ \mathrm{So},\\ \int {\mathrm{tan}}^{3}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{t}}^{2}.\mathrm{tan}2\mathrm{xsec}2\mathrm{x}\text{\hspace{0.17em}}.\frac{\mathrm{dt}}{2\mathrm{sec}2\mathrm{xtan}2\mathrm{x}}-\frac{1}{2}\mathrm{sec}2\mathrm{x}+\mathrm{C}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\text{\hspace{0.17em}}\int {\mathrm{t}}^{2}\text{\hspace{0.17em}}\mathrm{dt}-\frac{1}{2}\mathrm{sec}2\mathrm{x}+\mathrm{C}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\text{\hspace{0.17em}}×\frac{{\mathrm{t}}^{3}}{3}-\frac{1}{2}\mathrm{sec}2\mathrm{x}+\mathrm{C}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}\text{\hspace{0.17em}}{\mathrm{sec}}^{3}2\mathrm{x}-\frac{1}{2}\mathrm{sec}2\mathrm{x}+\mathrm{C}\end{array}$

Q.3 Find the integral of the function tan4x.

Ans.

$\begin{array}{l}\int {\mathrm{tan}}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{tan}}^{2}{\mathrm{xtan}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \left({\mathrm{sec}}^{2}\mathrm{x}-1\right){\mathrm{tan}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{sec}}^{2}\mathrm{x}{\mathrm{tan}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int {\mathrm{tan}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{sec}}^{2}\mathrm{x}{\mathrm{tan}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left({\mathrm{sec}}^{2}\mathrm{x}-1\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{sec}}^{2}\mathrm{x}{\mathrm{tan}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int {\mathrm{sec}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}+\int 1\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}=\mathrm{tanx}⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{sec}}^{2}\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{sec}}^{2}\mathrm{x}}\\ \int {\mathrm{tan}}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{sec}}^{2}\mathrm{x}\text{\hspace{0.17em}}.{\mathrm{t}}^{2}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{{\mathrm{sec}}^{2}\mathrm{x}}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{t}}^{2}\mathrm{dt}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{t}}^{3}}{3}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{{\mathrm{tan}}^{3}\mathrm{x}}{3}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\end{array}$

Q.4

$\mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{sin}}^{3}\mathrm{x}+{\mathrm{cos}}^{3}\mathrm{x}}{{\mathrm{sin}}^{2}{\mathrm{xcos}}^{2}\mathrm{x}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{sin}}^{\text{3}}{\text{x+cos}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}=\frac{{\text{sin}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}+\frac{{\text{cos}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{sinx}}{{\text{cos}}^{\text{2}}\text{x}}+\frac{\text{cosx}}{{\text{sin}}^{\text{2}}\text{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tanxsecx}+\mathrm{cotxcosecx}\\ \therefore \int \frac{{\text{sin}}^{\text{3}}{\text{x+cos}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}\text{\hspace{0.17em}}\mathrm{dx}=\int \mathrm{tanxsecx}\text{\hspace{0.17em}}\mathrm{dx}+\int \mathrm{cotxcosecx}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{secx}-\mathrm{cosecx}+\mathrm{C}\end{array}$

Q.5

$\mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{\mathrm{cos}2\mathrm{x}+2{\mathrm{sin}}^{2}\mathrm{x}}{{\mathrm{cos}}^{2}\mathrm{x}}$ $\begin{array}{l}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{cos}2\mathrm{x}+2{\mathrm{sin}}^{2}\mathrm{x}}{{\mathrm{cos}}^{2}\mathrm{x}}=\frac{\mathrm{cos}2\mathrm{x}+\left(1-\mathrm{cos}2\mathrm{x}\right)}{{\mathrm{cos}}^{2}\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{{\mathrm{cos}}^{2}\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sec}}^{2}\mathrm{x}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\int \frac{\mathrm{cos}2\mathrm{x}+2{\mathrm{sin}}^{2}\mathrm{x}}{{\mathrm{cos}}^{2}\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{sec}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tanx}+\mathrm{C}\end{array}$

Q.6

$\text{Find the integral of the function\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{sinxcos}}^{3}\mathrm{x}}.$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{sinx}\text{\hspace{0.17em}}{\mathrm{cos}}^{3}\mathrm{x}}=\frac{{\mathrm{sin}}^{2}\mathrm{x}+{\mathrm{cos}}^{2}\mathrm{x}}{{\mathrm{sinxcos}}^{3}\mathrm{x}}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{2}\mathrm{x}}{{\mathrm{sinxcos}}^{3}\mathrm{x}}+\frac{{\mathrm{cos}}^{2}\mathrm{x}}{{\mathrm{sinxcos}}^{3}\mathrm{x}}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tanx}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{sec}}^{2}\mathrm{x}+\frac{1}{\mathrm{sinx}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cosx}}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tanx}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{sec}}^{2}\mathrm{x}+\frac{{\mathrm{sec}}^{2}\mathrm{x}}{\mathrm{tanx}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int \frac{1}{{\mathrm{sinxcos}}^{3}\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=\int \mathrm{tanx}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{sec}}^{2}\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}+\int \frac{{\mathrm{sec}}^{2}\mathrm{x}}{\mathrm{tanx}}\text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}=\text{tanx}⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{sec}}^{2}\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{sec}}^{2}\mathrm{x}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int \frac{1}{\mathrm{sinx}\text{\hspace{0.17em}}{\mathrm{cos}}^{3}\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=\int \mathrm{t}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{sec}}^{2}\mathrm{x}\text{\hspace{0.17em}}.\frac{\mathrm{dt}}{{\mathrm{sec}}^{2}\mathrm{x}}+\int \frac{{\mathrm{sec}}^{2}\mathrm{x}}{\mathrm{t}}\text{\hspace{0.17em}}.\frac{\mathrm{dt}}{{\mathrm{sec}}^{2}\mathrm{x}}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \mathrm{t}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dt}+\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dt}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{t}}^{2}}{2}+\mathrm{log}\left|\mathrm{t}\right|+\mathrm{C}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int \frac{1}{{\mathrm{sinxcos}}^{3}\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\text{\hspace{0.17em}}=\frac{{\mathrm{tan}}^{2}\mathrm{x}}{2}+\mathrm{log}\left|\mathrm{tanx}\right|+\mathrm{C}\end{array}$

Q.7

$\mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{\mathrm{cos}2\mathrm{x}}{{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^{2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{cos}2\mathrm{x}}{{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^{2}}=\frac{\mathrm{cos}2\mathrm{x}}{\left({\mathrm{cos}}^{2}\mathrm{x}+{\mathrm{sin}}^{2}\mathrm{x}+2\mathrm{sinxcosx}\right)}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cos}2\mathrm{x}}{\left(1+\mathrm{sin}2\mathrm{x}\right)}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int \frac{\mathrm{cos}2\mathrm{x}}{{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\mathrm{cos}2\mathrm{x}}{\left(1+\mathrm{sin}2\mathrm{x}\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=1+\mathrm{sin}2\mathrm{x}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{cos}2\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{cos}2\mathrm{x}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int \frac{\mathrm{cos}2\mathrm{x}}{{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\mathrm{cos}2\mathrm{x}}{\mathrm{t}}\text{\hspace{0.17em}}.\frac{\mathrm{dt}}{2\mathrm{cos}2\mathrm{x}}\\ \text{ \hspace{0.17em}}=\frac{1}{2}\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{ \hspace{0.17em}}=\frac{1}{2}\mathrm{log}\left|\mathrm{t}\right|+\mathrm{C}\\ \text{ \hspace{0.17em}}=\frac{1}{2}\mathrm{log}\left|1+\mathrm{sin}2\mathrm{x}\right|+\mathrm{C}\\ \text{ \hspace{0.17em}}=\frac{1}{2}\mathrm{log}\left|{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^{2}\right|+\mathrm{C}\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×2\mathrm{log}\left(\mathrm{cosx}+\mathrm{sinx}\right)+\mathrm{C}\\ \text{Hence},\int \frac{\mathrm{cos}2\mathrm{x}}{{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\text{\hspace{0.17em}}=\mathrm{log}\left(\mathrm{cosx}+\mathrm{sinx}\right)+\mathrm{C}\end{array}$

Q.8

Find the integral of the function given below.
sin-1(cosx)

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \int {\mathrm{sin}}^{-1}\left(\mathrm{cosx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}=\text{cosx}⇒\mathrm{sinx}=\sqrt{1-{\mathrm{t}}^{2}}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dt}}{\mathrm{dx}}=-\mathrm{sinx}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{-\mathrm{sinx}}=\frac{-\mathrm{dt}}{\sqrt{1-{\mathrm{t}}^{2}}}\\ \therefore \int {\mathrm{sin}}^{-1}\left(\mathrm{cosx}\right)\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{sin}}^{-1}\mathrm{t}\text{\hspace{0.17em}}×\frac{-\mathrm{dt}}{\sqrt{1-{\mathrm{t}}^{2}}}\\ \text{ }=-\int \frac{{\mathrm{sin}}^{-1}\mathrm{t}}{\sqrt{1-{\mathrm{t}}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{ \hspace{0.17em}}=-\int \frac{\mathrm{u}}{\sqrt{1-{\mathrm{t}}^{2}}}\text{\hspace{0.17em}}\sqrt{1-{\mathrm{t}}^{2}}\mathrm{du}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Let}\text{u}={\mathrm{sin}}^{-1}\mathrm{t}\\ ⇒\frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{\sqrt{1-{\mathrm{t}}^{2}}}\\ \mathrm{dx}=\sqrt{1-{\mathrm{t}}^{2}}\mathrm{du}\end{array}\right]\\ \text{ }=-\int \mathrm{u}\text{\hspace{0.17em}}\mathrm{du}\\ \text{ }=-\frac{{\mathrm{u}}^{2}}{2}+\mathrm{C}\\ \text{ }=-\frac{{\left({\mathrm{sin}}^{-1}\mathrm{t}\right)}^{2}}{2}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int {\mathrm{sin}}^{-1}\left(\mathrm{cosx}\right)\text{\hspace{0.17em}}\mathrm{dx}\text{\hspace{0.17em}}=-\frac{{\left\{{\mathrm{sin}}^{-1}\left(\text{cosx}\right)\right\}}^{2}}{2}+\mathrm{C}\\ \text{ }=-\frac{{\left\{\frac{\mathrm{\pi }}{2}-{\mathrm{cos}}^{-1}\left(\text{cosx}\right)\right\}}^{2}}{2}+\mathrm{C}\text{ }\left[{\mathrm{sin}}^{-1}\mathrm{x}+{\mathrm{cos}}^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\right]\\ \text{ }=-\frac{{\left\{\frac{\mathrm{\pi }}{2}-\mathrm{x}\right\}}^{2}}{2}+\mathrm{C}\\ \text{ }=-\frac{1}{2}\left(\frac{{\mathrm{\pi }}^{2}}{4}-\mathrm{\pi x}+{\mathrm{x}}^{2}\right)+\mathrm{C}\\ \text{ }=-\frac{{\mathrm{\pi }}^{2}}{8}+\frac{\mathrm{\pi x}}{2}-\frac{{\mathrm{x}}^{2}}{2}+\mathrm{C}\\ \text{ }=-\frac{{\mathrm{x}}^{2}}{2}+\frac{\mathrm{\pi x}}{2}+\left(\mathrm{C}-\frac{{\mathrm{\pi }}^{2}}{8}\right)\\ \text{ }=\frac{\mathrm{\pi x}}{2}-\frac{{\mathrm{x}}^{2}}{2}+\mathrm{C}‘,\text{where C’=}\mathrm{C}+\frac{{\mathrm{\pi }}^{2}}{8}\end{array}$

Q.9

$\mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have}\frac{1}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left[\frac{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\right]\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left[\frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{b}-\mathrm{x}+\mathrm{a}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\right]\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left[\frac{\mathrm{sin}\left\{\left(\mathrm{x}-\mathrm{b}\right)-\left(\mathrm{x}-\mathrm{a}\right)\right\}}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\right]\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left\{\frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)-\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{sin}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\right\}\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left\{\frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}-\frac{\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)\mathrm{sin}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\right\}\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left\{\frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}-\frac{\mathrm{sin}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)}\right\}\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left\{\mathrm{tan}\left(\mathrm{x}-\mathrm{b}\right)-\mathrm{tan}\left(\mathrm{x}-\mathrm{a}\right)\right\}\\ \therefore \int \frac{1}{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\text{ }\left\{\int \mathrm{tan}\left(\mathrm{x}-\mathrm{b}\right)\text{\hspace{0.17em}}\mathrm{dx}\text{ }-\int \mathrm{tan}\left(\mathrm{x}-\mathrm{a}\right)\text{\hspace{0.17em}}\mathrm{dx}\right\}\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\left\{-\mathrm{log}\left|\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)\right|+\mathrm{log}\left|\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)\right|\right\}+\mathrm{C}\\ \text{ }=\frac{1}{\mathrm{sin}\left(\mathrm{a}-\mathrm{b}\right)}\mathrm{log}\left|\frac{\mathrm{cos}\left(\mathrm{x}-\mathrm{a}\right)}{\mathrm{cos}\left(\mathrm{x}-\mathrm{b}\right)}\right|+\mathrm{C}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{{\mathrm{sin}}^{2}\text{x}-{\mathrm{cos}}^{2}\text{x}}{{\mathrm{sin}}^{2}{\mathrm{xcos}}^{2}\text{x}}\text{ }\mathrm{dx}\text{\hspace{0.17em}is equal to}\\ \left(\text{A}\right)\text{ }\mathrm{tanx}+\mathrm{cotx}+\text{C}\\ \left(\text{B}\right)\text{ }\mathrm{tanx}+\mathrm{cosec}\text{x}+\text{C}\\ \left(\text{C}\right)\text{ }-\mathrm{tanx}+\mathrm{cotx}+\text{C}\\ \left(\text{D}\right)\text{ }\mathrm{tanx}+\mathrm{secx}+\text{C}\end{array}$

Ans.

$\begin{array}{l}\int \frac{{\mathrm{sin}}^{2}\mathrm{x}-{\mathrm{cos}}^{2}\mathrm{x}}{{\mathrm{sin}}^{2}{\mathrm{xcos}}^{2}\mathrm{x}}\mathrm{ }\mathrm{dx}=\int \frac{{\mathrm{sin}}^{2}\mathrm{x}}{{\mathrm{sin}}^{2}{\mathrm{xcos}}^{2}\mathrm{x}}\mathrm{ }\mathrm{dx}-\int \frac{{\mathrm{cos}}^{2}\mathrm{x}}{{\mathrm{sin}}^{2}{\mathrm{xcos}}^{2}\mathrm{x}}\mathrm{ }\mathrm{dx}\\ \mathrm{ } \mathrm{ }=\int {\mathrm{sec}}^{2}\mathrm{x}\mathrm{ }\mathrm{dx}-\int {\mathrm{cosec}}^{2}\mathrm{x}\mathrm{ }\mathrm{dx}\\ \mathrm{ } \mathrm{ }=\mathrm{tanx}+\mathrm{cotx}+\mathrm{C}\\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{{\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)}{{\mathrm{cos}}^{2}\left({\mathrm{e}}^{\mathrm{x}}\mathrm{x}\right)}\text{ }\mathrm{dx}\text{ }\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }-\mathrm{cot}\left({\mathrm{ex}}^{\mathrm{x}}\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\mathrm{tan}\left({\mathrm{xe}}^{\mathrm{x}}\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }\mathrm{tan}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\mathrm{cot}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have}\int \frac{{\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)}{{\mathrm{cos}}^{2}\left({\mathrm{e}}^{\mathrm{x}}\mathrm{x}\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}={\mathrm{e}}^{\mathrm{x}}\mathrm{x}\\ ⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}+\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}+{\mathrm{xe}}^{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)}\\ \therefore \int \frac{{\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)}{{\mathrm{cos}}^{2}\left({\mathrm{e}}^{\mathrm{x}}\mathrm{x}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)}{{\mathrm{cos}}^{2}\mathrm{t}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{{\mathrm{cos}}^{2}\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{sec}}^{2}\mathrm{t}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tant}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{tan}\left({\mathrm{xe}}^{\mathrm{x}}\right)+\mathrm{C}\\ \mathrm{Hence},\text{the correct option is B.}\end{array}$

Q.12

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{3{\mathrm{x}}^{2}}{{\mathrm{x}}^{6}+1}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{3}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^{2}\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{3{\mathrm{x}}^{2}}\\ \therefore \int \frac{3{\mathrm{x}}^{2}}{{\mathrm{x}}^{6}+1}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{3{\mathrm{x}}^{2}}{{\left({\mathrm{x}}^{3}\right)}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{3{\mathrm{x}}^{2}}{{\mathrm{t}}^{2}+1}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{3{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{{\mathrm{t}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{tan}}^{-1}\mathrm{t}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{tan}}^{-1}\left({\mathrm{x}}^{3}\right)+\mathrm{C}\end{array}$

Q.13

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{1}{\sqrt{1+4{\mathrm{x}}^{2}}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{1+4{\mathrm{x}}^{2}}}=\frac{1}{\sqrt{1+{\left(2\mathrm{x}\right)}^{2}}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=2\mathrm{x}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{2}\\ \therefore \int \frac{1}{\sqrt{1+4{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\sqrt{1+{\left(2\mathrm{x}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{1+{\mathrm{t}}^{2}}}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\int \frac{1}{\sqrt{1+{\mathrm{t}}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}+1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|2\mathrm{x}+\sqrt{{\left(2\mathrm{x}\right)}^{2}+1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|2\mathrm{x}+\sqrt{4{\mathrm{x}}^{2}+1}|+\mathrm{C}\end{array}$

Q.14

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{1}{\sqrt{{\left(2-\mathrm{x}\right)}^{2}+1}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{{\left(2-\mathrm{x}\right)}^{2}+1}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=2-\mathrm{x}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=-1\\ ⇒\mathrm{dx}=-\text{\hspace{0.17em}}\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{{\left(2-\mathrm{x}\right)}^{2}+1}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\sqrt{{\mathrm{t}}^{2}+1}}\text{\hspace{0.17em}}\left(-\mathrm{dt}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\int \frac{1}{\sqrt{1+{\mathrm{t}}^{2}}}.\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}+1}|+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\because \text{\hspace{0.17em}\hspace{0.17em}}\int \frac{1}{\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}|\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|\left(2-\mathrm{x}\right)+\sqrt{{\left(2-\mathrm{x}\right)}^{2}+1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{1}{\left(2-\mathrm{x}\right)+\sqrt{{\left(2-\mathrm{x}\right)}^{2}+1}}|+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because -\mathrm{logx}=\mathrm{log}\left(\frac{1}{\mathrm{x}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{1}{\left(2-\mathrm{x}\right)+\sqrt{4-4\mathrm{x}+{\mathrm{x}}^{2}+1}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{1}{\left(2-\mathrm{x}\right)+\sqrt{{\mathrm{x}}^{2}-4\mathrm{x}+5}}|+\mathrm{C}\end{array}$

Q.15

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}.\phantom{\rule{0ex}{0ex}}\frac{1}{\sqrt{9-25{\mathrm{x}}^{2}}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{ have,}\frac{1}{\sqrt{9-25{\mathrm{x}}^{2}}}=\frac{1}{\sqrt{9-{\left(5\mathrm{x}\right)}^{2}}}\\ \mathrm{Let}\text{\hspace{0.17em} }\mathrm{t}=5\mathrm{x}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=5\\ ⇒\mathrm{dx}=\frac{1}{5}\text{\hspace{0.17em}}\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{9-25{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{5}\int \frac{1}{\sqrt{9-{\left(5\mathrm{x}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{ }=\frac{1}{5}\int \frac{1}{\sqrt{{3}^{2}-{\mathrm{t}}^{2}}}.\text{\hspace{0.17em}}\mathrm{dt}\\ \text{ \hspace{0.17em}}=\frac{1}{5}{\mathrm{sin}}^{-1}\left(\frac{\mathrm{t}}{3}\right)+\mathrm{C}\\ \text{ }=\frac{1}{5}{\mathrm{sin}}^{-1}\left(\frac{5\mathrm{x}}{3}\right)+\mathrm{C}\end{array}$

Q.16

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{3\mathrm{x}}{1+2{\mathrm{x}}^{4}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{3\mathrm{x}}{1+2{\mathrm{x}}^{4}}=\frac{3\mathrm{x}}{1+{\left(\sqrt{2}{\mathrm{x}}^{2}\right)}^{2}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\sqrt{2}{\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\sqrt{2}\mathrm{x}\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\sqrt{2}\mathrm{x}}\\ \therefore \int \frac{3\mathrm{x}}{1+2{\mathrm{x}}^{4}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{3\mathrm{x}}{1+{\left(\sqrt{2}{\mathrm{x}}^{2}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{3\mathrm{x}}{1+{\mathrm{t}}^{2}}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{2\sqrt{2}\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2\sqrt{2}}\int \frac{1}{1+{\mathrm{t}}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2\sqrt{2}}{\mathrm{tan}}^{-1}\mathrm{t}+\mathrm{C}\text{\hspace{0.17em}}=\frac{3}{2\sqrt{2}}{\mathrm{tan}}^{-1}\left(\sqrt{2}{\mathrm{x}}^{2}\right)+\mathrm{C}\end{array}$

Q.17

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{{\mathrm{x}}^{2}}{1-{\mathrm{x}}^{6}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{{\mathrm{x}}^{2}}{1-{\mathrm{x}}^{6}}=\frac{{\mathrm{x}}^{2}}{1-{\left({\mathrm{x}}^{3}\right)}^{2}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{3}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^{2}\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{3{\mathrm{x}}^{2}}\\ \therefore \int \frac{{\mathrm{x}}^{2}}{1-{\mathrm{x}}^{6}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{x}}^{2}}{1-{\left({\mathrm{x}}^{3}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{{\mathrm{x}}^{2}}{1-{\mathrm{t}}^{2}}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{3{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\int \frac{1}{1-{\mathrm{t}}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\left(\frac{1}{2}\mathrm{log}|\frac{1+\mathrm{t}}{1-\mathrm{t}}|\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}\mathrm{log}|\frac{1+{\mathrm{x}}^{3}}{1-{\mathrm{x}}^{3}}|+\mathrm{C}\end{array}$

Q.18

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^{2}-1}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^{2}-1}}=\frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^{2}-1}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{2}-1⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\\ \therefore \int \frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{1}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{\mathrm{x}}{\sqrt{\mathrm{t}}}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{2\mathrm{x}}-\int \frac{1}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\int {\mathrm{t}}^{-\frac{1}{2}}.\text{\hspace{0.17em}}\mathrm{dt}-\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(2\sqrt{\mathrm{t}}\right)-\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{\mathrm{t}}-\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-1}-\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-1}|+\mathrm{C}\end{array}$

Q.19

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{{\mathrm{x}}^{2}}{\sqrt{{\mathrm{x}}^{6}+{\mathrm{a}}^{6}}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{{\mathrm{x}}^{2}}{\sqrt{{\mathrm{x}}^{6}+{\mathrm{a}}^{6}}}=\frac{{\mathrm{x}}^{2}}{\sqrt{{\mathrm{a}}^{6}+{\left({\mathrm{x}}^{3}\right)}^{2}}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{3}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^{2}\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{3{\mathrm{x}}^{2}}\\ \therefore \int \frac{{\mathrm{x}}^{2}}{\sqrt{{\mathrm{x}}^{6}+{\mathrm{a}}^{6}}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{x}}^{2}}{\sqrt{{\left({\mathrm{a}}^{3}\right)}^{2}+{\left({\mathrm{x}}^{3}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \frac{{\mathrm{x}}^{2}}{\sqrt{{\left({\mathrm{a}}^{3}\right)}^{2}+{\mathrm{t}}^{2}}}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{3{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\int \frac{1}{\sqrt{{\left({\mathrm{a}}^{3}\right)}^{2}+{\mathrm{t}}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}+{\left({\mathrm{a}}^{3}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{3}\mathrm{log}|{\mathrm{x}}^{3}+\sqrt{{\left({\mathrm{x}}^{3}\right)}^{2}+{\left({\mathrm{a}}^{3}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\mathrm{log}|{\mathrm{x}}^{3}+\sqrt{{\mathrm{x}}^{6}+{\mathrm{a}}^{6}}|+\mathrm{C}\end{array}$

Q.20

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{{\mathrm{sec}}^{2}\mathrm{x}}{\sqrt{{\mathrm{tan}}^{2}\mathrm{x}+4}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{{\mathrm{sec}}^{2}\mathrm{x}}{\sqrt{{\mathrm{tan}}^{2}\mathrm{x}+4}}=\frac{{\mathrm{sec}}^{2}\mathrm{x}}{\sqrt{{\mathrm{tan}}^{2}\mathrm{x}+{2}^{2}}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\mathrm{tanx}⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{sec}}^{2}\mathrm{x}\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{sec}}^{2}\mathrm{x}}\\ \therefore \int \frac{{\mathrm{sec}}^{2}\mathrm{x}}{\sqrt{{\mathrm{tan}}^{2}\mathrm{x}+4}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{sec}}^{2}\mathrm{x}}{\sqrt{{\mathrm{tan}}^{2}\mathrm{x}+{2}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{{\mathrm{sec}}^{2}\mathrm{x}}{\sqrt{{\mathrm{t}}^{2}+{2}^{2}}}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{{\mathrm{sec}}^{2}\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{{\mathrm{t}}^{2}+{2}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}+{\left(2\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\mathrm{tanx}+\sqrt{{\mathrm{tan}}^{2}\mathrm{x}+4}|+\mathrm{C}\end{array}$

Q.21

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \frac{1}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+2}}=\frac{1}{\sqrt{{\left(\mathrm{x}+1\right)}^{2}+1}}\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{t}=\mathrm{x}+1⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\\ ⇒\mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\sqrt{{\left(\mathrm{x}+1\right)}^{2}+1}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{{\mathrm{t}}^{2}+{1}^{2}}}.\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}+{\left(1\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\left(\mathrm{x}+1\right)+\sqrt{{\left(\mathrm{x}+1\right)}^{2}+1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\left(\mathrm{x}+1\right)+\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+2}|+\mathrm{C}\end{array}$

Q.22

$\text{Integrate the function\hspace{0.17em}}\frac{1}{9{\mathrm{x}}^{2}+6\mathrm{x}+5}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{9{\mathrm{x}}^{2}+6\mathrm{x}+5}=\frac{1}{{\left(3\mathrm{x}+1\right)}^{2}+{2}^{2}}\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{t}=3\mathrm{x}+1⇒\frac{\mathrm{dt}}{\mathrm{dx}}=3\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{3}\\ \therefore \int \frac{1}{9{\mathrm{x}}^{2}+6\mathrm{x}+5}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{{\left(3\mathrm{x}+1\right)}^{2}+{2}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{{\mathrm{t}}^{2}+{2}^{2}}.\text{\hspace{0.17em}}\frac{\mathrm{dt}}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\int \frac{1}{{\mathrm{t}}^{2}+{2}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\left(\frac{1}{2}{\mathrm{tan}}^{-1}\frac{\mathrm{t}}{2}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}{\mathrm{tan}}^{-1}\frac{\left(3\mathrm{x}+1\right)}{2}+\mathrm{C}\end{array}$

Q.23

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{7-6\mathrm{x}-{\mathrm{x}}^{2}}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{7-6\mathrm{x}-{\mathrm{x}}^{2}}}=\frac{1}{\sqrt{16-{\left(\mathrm{x}+3\right)}^{2}}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\mathrm{x}+3⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\\ ⇒\mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{7-6\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\sqrt{16-{\left(\mathrm{x}+3\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{{4}^{2}-{\mathrm{t}}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{-1}\left(\frac{\mathrm{t}}{4}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}+3}{4}\right)+\mathrm{C}\end{array}$

Q.24

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}}=\frac{1}{\sqrt{{\mathrm{x}}^{2}-3\mathrm{x}+2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}-\frac{1}{4}}}\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{t}=\mathrm{x}-\frac{3}{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\\ ⇒\mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\sqrt{{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}-\frac{1}{4}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{{\mathrm{t}}^{2}-{\left(\frac{1}{2}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}-{\left(\frac{1}{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\left(\mathrm{x}-\frac{3}{2}\right)+\sqrt{{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\left(\mathrm{x}-\frac{3}{2}\right)+\sqrt{{\mathrm{x}}^{2}-3\mathrm{x}+2}|+\mathrm{C}\end{array}$

Q.25

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^{2}}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^{2}}}=\frac{1}{\sqrt{8-\left({\mathrm{x}}^{2}-3\mathrm{x}+\frac{9}{4}-\frac{9}{4}\right)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{8+\frac{9}{4}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{\frac{41}{4}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}}\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{t}=\mathrm{x}-\frac{3}{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\\ ⇒\mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\sqrt{\frac{41}{4}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{{\left(\frac{\sqrt{41}}{2}\right)}^{2}-{\mathrm{t}}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{-1}\left(\frac{\mathrm{t}}{\frac{\sqrt{41}}{2}}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{-1}\left(\frac{2\mathrm{t}}{\sqrt{41}}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{-1}\left\{\frac{2\left(\mathrm{x}-\frac{3}{2}\right)}{\sqrt{41}}\right\}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{-1}\left\{\frac{\left(2\mathrm{x}-3\right)}{\sqrt{41}}\right\}+\mathrm{C}\end{array}$

Q.26

$\text{Integrate the function}\frac{1}{\sqrt{\left(\mathrm{x}–\mathrm{a}\right)\left(\mathrm{x}–\mathrm{b}\right)}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{\left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{x}-\mathrm{b}\right)}}=\frac{1}{\sqrt{{\mathrm{x}}^{2}-\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{{\left(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}}}\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{t}=\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\\ ⇒\mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{\left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{x}-\mathrm{b}\right)}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\sqrt{{\left(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{{\mathrm{t}}^{2}-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{{\mathrm{t}}^{2}-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\left(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)+\sqrt{{\left(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\left(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)+\sqrt{{\mathrm{x}}^{2}-\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab}}|+\mathrm{C}\end{array}$

Q.27

$\text{Integrate the function\hspace{0.17em}}\frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^{2}+\mathrm{x}-3}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^{2}+\mathrm{x}-3}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}+1=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\mathrm{x}}^{2}+\mathrm{x}-3\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{A}\left(4\mathrm{x}+1\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 4=4\mathrm{A}\text{and A}+\mathrm{B}=1⇒\mathrm{A}=1\text{and}\mathrm{B}=1-1=0\\ \mathrm{Now},\text{​\hspace{0.17em}\hspace{0.17em}let t}=2{\mathrm{x}}^{2}+\mathrm{x}-3⇒\frac{\mathrm{dt}}{\mathrm{dx}}=4\mathrm{x}+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{4\mathrm{x}+1}\\ \therefore \int \frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^{2}+\mathrm{x}-3}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^{2}+\mathrm{x}-3}}\text{\hspace{0.17em}}.\frac{\mathrm{dt}}{4\mathrm{x}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}.\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{\mathrm{t}}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=2\sqrt{2{\mathrm{x}}^{2}+\mathrm{x}-3}+\mathrm{C}\end{array}$

Q.28

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}-1}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}-1}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}-1\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{A}\left(2\mathrm{x}\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 1=2\mathrm{A}\text{and}\mathrm{B}=2⇒\mathrm{A}=\frac{1}{2}\text{and}\mathrm{B}=2\\ \mathrm{Therefore},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+2=\frac{1}{2}\left(2\mathrm{x}\right)+2\\ \therefore \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\frac{1}{2}\left(2\mathrm{x}\right)+2}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{\frac{1}{2}\left(2\mathrm{x}\right)}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{2}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Now},\text{​\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}let t}={\mathrm{x}}^{2}-1⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\\ \mathrm{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{\left(2\mathrm{x}\right)}{\sqrt{\mathrm{t}}}.\frac{\mathrm{dt}}{2\mathrm{x}}+\int \frac{2}{\sqrt{{\mathrm{x}}^{2}-1}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}.\mathrm{dt}+2\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-1}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{\mathrm{t}}+2\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-1}+2\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-1}|+\mathrm{C}\end{array}$

Q.29

$\text{Integrate the function\hspace{0.17em}}\frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^{2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^{2}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}5\mathrm{x}-2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+2\mathrm{x}+3{\mathrm{x}}^{2}\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{A}\left(6\mathrm{x}+2\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5=6\mathrm{A}\text{and 2A}+\mathrm{B}=-2\\ ⇒\mathrm{A}=\frac{5}{6}\text{and}\mathrm{B}=-2-2\left(\frac{5}{6}\right)=-2-\frac{5}{3}=-\frac{11}{3}\\ \therefore \text{\hspace{0.17em}}5\mathrm{x}-2=\frac{5}{6}\left(6\mathrm{x}+2\right)-\frac{11}{3}\\ \therefore \int \frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\frac{5}{6}\left(6\mathrm{x}+2\right)-\frac{11}{3}}{\sqrt{2{\mathrm{x}}^{2}+\mathrm{x}-3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{5}{6}\int \frac{\left(6\mathrm{x}+2\right)}{\sqrt{1+2\mathrm{x}+3{\mathrm{x}}^{2}}}\mathrm{dx}-\frac{11}{3}\int \frac{1}{\sqrt{1+2\mathrm{x}+3{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Now},\text{​\hspace{0.17em}\hspace{0.17em}let t}=1+2\mathrm{x}+3{\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=6\mathrm{x}+2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{6\mathrm{x}+2},\text{​\hspace{0.17em}}\mathrm{then}\\ \int \frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{5}{6}\int \frac{\left(6\mathrm{x}+2\right)}{1+2\mathrm{x}+3{\mathrm{x}}^{2}}\text{\hspace{0.17em}}.\frac{\mathrm{dt}}{6\mathrm{x}+2}-\frac{11}{3}\int \frac{1}{1+2\mathrm{x}+3{\mathrm{x}}^{2}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{6}\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}}.\mathrm{dt}-\frac{11}{3}\int \frac{1}{3\left\{{\left(\mathrm{x}+\frac{1}{3}\right)}^{2}+{\left(\frac{\sqrt{2}}{3}\right)}^{2}\right\}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{5}{6}\mathrm{log}|\mathrm{t}|-\frac{11}{9}\int \frac{1}{{\left(\mathrm{x}+\frac{1}{3}\right)}^{2}+{\left(\frac{\sqrt{2}}{3}\right)}^{2}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{6}\mathrm{log}|1+2\mathrm{x}+3{\mathrm{x}}^{2}|-\frac{11}{9}×\frac{1}{\left(\frac{\sqrt{2}}{3}\right)}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{6}\mathrm{log}|1+2\mathrm{x}+3{\mathrm{x}}^{2}|-\frac{11}{3\sqrt{2}}{\mathrm{tan}}^{-1}\left(\frac{3\mathrm{x}+1}{\sqrt{2}}\right)+\mathrm{C}\end{array}$

Q.30

$\text{Integrate the function\hspace{0.17em}}\frac{6\mathrm{x}+7}{\sqrt{\left(\mathrm{x}-5\right)\left(\mathrm{x}-4\right)}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{6\mathrm{x}+7}{\sqrt{\left(\mathrm{x}-5\right)\left(\mathrm{x}-4\right)}}=\frac{6\mathrm{x}+7}{\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}6\mathrm{x}+7=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}-9\mathrm{x}+20\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{x}+7=\mathrm{A}\left(2\mathrm{x}-9\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6=2\mathrm{A}\text{and}-\text{9A}+\mathrm{B}=7\\ ⇒\mathrm{A}=\frac{6}{2}=\text{3 and}\mathrm{B}=7+9\left(3\right)=34\\ \therefore \text{\hspace{0.17em}}6\mathrm{x}+7=3\left(2\mathrm{x}-9\right)+34\\ \therefore \int \frac{6\mathrm{x}+7}{\sqrt{\left(\mathrm{x}-5\right)\left(\mathrm{x}-4\right)}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{6\mathrm{x}+7}{\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{3\left(2\mathrm{x}-9\right)+34}{\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\int \frac{\left(2\mathrm{x}-9\right)}{\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{34}{\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Now},\text{​\hspace{0.17em}\hspace{0.17em}let t}={\mathrm{x}}^{2}-9\mathrm{x}+20⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}-9\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}-9},\text{​\hspace{0.17em}}\mathrm{then}\\ \int \frac{6\mathrm{x}+7}{\sqrt{\left(\mathrm{x}-5\right)\left(\mathrm{x}-4\right)}}\text{\hspace{0.17em}}\mathrm{dx}=3\int \frac{\left(2\mathrm{x}-9\right)}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}.\frac{\mathrm{dt}}{2\mathrm{x}-9}+34\int \frac{1}{\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\int \frac{1}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}.\mathrm{dt}+34\int \frac{1}{\sqrt{\left\{{\left(\mathrm{x}-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}\right\}}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3×2\sqrt{\mathrm{t}}+34\mathrm{log}\left|\left(\mathrm{x}-\frac{9}{2}\right)+\sqrt{\left\{{\left(\mathrm{x}-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}\right\}}\right|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}+34\mathrm{log}\left|\left(\mathrm{x}-\frac{9}{2}\right)+\sqrt{{\mathrm{x}}^{2}-9\mathrm{x}+20}\right|+\mathrm{C}\end{array}$

Q.31

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(4\mathrm{x}-{\mathrm{x}}^{2}\right)+\mathrm{B}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{x}+2=\mathrm{A}\left(4-2\mathrm{x}\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=-2\mathrm{A}\text{and 4A}+\mathrm{B}=2\\ ⇒\mathrm{A}=-\frac{1}{2}\text{and}\mathrm{B}=2-4\left(-\frac{1}{2}\right)=4\\ \therefore \text{\hspace{0.17em}}\mathrm{x}+2=-\frac{1}{2}\left(4-2\mathrm{x}\right)+4\\ \therefore \int \frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{-\frac{1}{2}\left(4-2\mathrm{x}\right)+4}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{2}\int \frac{\left(4-2\mathrm{x}\right)}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{4}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\int \frac{\left(4-2\mathrm{x}\right)}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{4}{\sqrt{4-\left(4-4\mathrm{x}+{\mathrm{x}}^{2}\right)}}\text{\hspace{0.17em}}.\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{2}\int \frac{\left(4-2\mathrm{x}\right)}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\mathrm{dx}+\int \frac{4}{\sqrt{{2}^{2}-{\left(2-\mathrm{x}\right)}^{2}}}\mathrm{dx}\\ \mathrm{Now},\text{​\hspace{0.17em}\hspace{0.17em}let t}=4\mathrm{x}-{\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=4-2\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{4-2\mathrm{x}},\text{​\hspace{0.17em}}\mathrm{then}\\ \int \frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}=-\frac{1}{2}\int \frac{\left(4-2\mathrm{x}\right)}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{4-2\mathrm{x}}+\int \frac{4}{\sqrt{{2}^{2}-{\left(2-\mathrm{x}\right)}^{2}}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{2}×2\sqrt{\mathrm{t}}-4{\mathrm{sin}}^{-1}\left(\frac{2-\mathrm{x}}{2}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\sqrt{4\mathrm{x}-{\mathrm{x}}^{2}}-4{\mathrm{sin}}^{-1}\left(\frac{2-\mathrm{x}}{2}\right)+\mathrm{C}\end{array}$

Q.32

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\\ \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{2\left(\mathrm{x}+2\right)}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\int \frac{2\mathrm{x}+4}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\int \frac{2\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{2}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}={\mathrm{x}}^{2}+2\mathrm{x}+3⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}+2\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}+2},\mathrm{then}\\ \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{2\mathrm{x}+2}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{2\mathrm{x}+2}+\frac{1}{2}\int \frac{2}{\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\int \frac{1}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}\mathrm{dt}+\int \frac{1}{\sqrt{{\left(\mathrm{x}+1\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}.2\sqrt{\mathrm{t}}+\mathrm{log}|\left(\mathrm{x}+1\right)+\sqrt{{\left(\mathrm{x}+1\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}+\mathrm{log}|\left(\mathrm{x}+1\right)+\sqrt{{\mathrm{x}}^{2}+2\mathrm{x}+3}|+\mathrm{C}\end{array}$

Q.33

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+3}{{\mathrm{x}}^{2}-2\mathrm{x}-5}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}+3}{{\mathrm{x}}^{2}-2\mathrm{x}-5}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+3=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}-2\mathrm{x}-5\right)+\mathrm{B}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{x}+3=\mathrm{A}\left(2\mathrm{x}-2\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=2\mathrm{A}\text{and}-\text{2A}+\mathrm{B}=3\\ ⇒\mathrm{A}=\frac{1}{2}\text{and}\mathrm{B}=3+2\left(\frac{1}{2}\right)=4\\ \therefore \text{\hspace{0.17em}}\mathrm{x}+3=\frac{1}{2}\left(2\mathrm{x}-2\right)+4\\ ⇒\int \frac{\mathrm{x}+3}{{\mathrm{x}}^{2}-2\mathrm{x}-5}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\frac{1}{2}\left(2\mathrm{x}-2\right)+4}{{\mathrm{x}}^{2}-2\mathrm{x}-5}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\int \frac{\left(2\mathrm{x}-2\right)}{{\mathrm{x}}^{2}-2\mathrm{x}-5}\text{\hspace{0.17em}}\mathrm{dx}+4\int \frac{1}{{\mathrm{x}}^{2}-2\mathrm{x}-5}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}={\mathrm{x}}^{2}-2\mathrm{x}-5⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}-2\\ ⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}-2},\mathrm{then}\\ ⇒\int \frac{\mathrm{x}+3}{{\mathrm{x}}^{2}-2\mathrm{x}-5}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{\left(2\mathrm{x}-2\right)}{\mathrm{t}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{2\mathrm{x}-2}+4\int \frac{1}{{\left(\mathrm{x}-1\right)}^{2}-6}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}+4\int \frac{1}{{\left(\mathrm{x}-1\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\mathrm{t}|+4×\frac{1}{2\sqrt{6}}\mathrm{log}\left(\frac{\mathrm{x}-1-\sqrt{6}}{\mathrm{x}-1+\sqrt{6}}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\mathrm{log}|{\mathrm{x}}^{2}-2\mathrm{x}-5|+\frac{2}{\sqrt{6}}\mathrm{log}\left(\frac{\mathrm{x}-1-\sqrt{6}}{\mathrm{x}-1+\sqrt{6}}\right)+\mathrm{C}\end{array}$

Q.34

$\text{Integrate the function\hspace{0.17em}}\frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}5\mathrm{x}+3=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}+4\mathrm{x}+10\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}5\mathrm{x}+3=\mathrm{A}\left(2\mathrm{x}+4\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}5=2\mathrm{A}\text{and 4A}+\mathrm{B}=3\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\frac{5}{2}\text{and}\mathrm{B}=3-4\left(\frac{5}{2}\right)=-7\\ \therefore \text{\hspace{0.17em}}5\mathrm{x}+3=\frac{5}{2}\left(2\mathrm{x}+4\right)-7\\ \therefore \int \frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\frac{5}{2}\left(2\mathrm{x}+4\right)-7}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}}=\frac{5}{2}\int \frac{\left(2\mathrm{x}+4\right)}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{7}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{5}{2}\int \frac{\left(2\mathrm{x}+4\right)}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{7}{\sqrt{{\left(\mathrm{x}+2\right)}^{2}+6}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{5}{2}\int \frac{\left(2\mathrm{x}+4\right)}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\text{\hspace{0.17em}}\mathrm{dx}-7\int \frac{1}{\sqrt{{\left(\mathrm{x}+2\right)}^{2}+{\left(\sqrt{6}\right)}^{2}}}\mathrm{dx}\\ \mathrm{Now},\text{​\hspace{0.17em}\hspace{0.17em}let t}={\mathrm{x}}^{2}+4\mathrm{x}+10⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}+4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}+4},\text{​\hspace{0.17em}}\mathrm{then}\\ \int \frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{5}{2}\int \frac{\left(2\mathrm{x}+4\right)}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{2\mathrm{x}+4}-7\int \frac{1}{\sqrt{{\left(\mathrm{x}+2\right)}^{2}+{\left(\sqrt{6}\right)}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{5}{2}×2\sqrt{\mathrm{t}}-7\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\left(\mathrm{x}+2\right)}^{2}+{\left(\sqrt{6}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=5\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}-7\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+10}|+\mathrm{C}\end{array}$

Q.35

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{\mathrm{dx}}{{\mathrm{x}}^{2}+2\mathrm{x}+2}\\ \left(\mathrm{A}\right)\text{​ }\mathrm{x}\text{ }{\mathrm{tan}}^{-1}\left(\mathrm{x}+1\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }{\mathrm{tan}}^{-1}\left(\mathrm{x}+1\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }\left(\mathrm{x}+1\right){\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \int \frac{\mathrm{dx}}{{\mathrm{x}}^{2}+2\mathrm{x}+2}=\int \frac{\mathrm{dx}}{{\left(\mathrm{x}+1\right)}^{2}+{1}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}+1}{1}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{tan}}^{-1}\left(\mathrm{x}+1\right)+\mathrm{C}\\ \mathrm{Hence},\text{​ the correct option is B.}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{\mathrm{dx}}{\sqrt{9\mathrm{x}-4{\mathrm{x}}^{2}}}\text{ }\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }\frac{1}{9}\text{ }{\mathrm{sin}}^{-1}\left(\frac{9\mathrm{x}-8}{8}\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\frac{1}{2}\text{ }{\mathrm{sin}}^{-1}\left(\frac{8\mathrm{x}-9}{9}\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }\frac{1}{3}\text{ }{\mathrm{sin}}^{-1}\left(\frac{9\mathrm{x}-8}{8}\right)+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\frac{1}{2}\text{ }{\mathrm{sin}}^{-1}\left(\frac{9\mathrm{x}-8}{9}\right)+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\int \frac{\mathrm{dx}}{\sqrt{9\mathrm{x}-4{\mathrm{x}}^{2}}}=\int \frac{\mathrm{dx}}{\sqrt{-4\left({\mathrm{x}}^{2}-\frac{9}{4}\mathrm{x}\right)}}\\ =\int \frac{\mathrm{dx}}{\sqrt{-4\left\{{\mathrm{x}}^{2}-\frac{9}{4}\mathrm{x}+{\left(\frac{9}{8}\right)}^{2}-{\left(\frac{9}{8}\right)}^{2}\right\}}}\\ =\frac{1}{2}\int \frac{\mathrm{dx}}{\sqrt{{\left(\frac{9}{8}\right)}^{2}-{\left(\mathrm{x}-\frac{9}{8}\right)}^{2}}}\\ =\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}-\frac{9}{8}}{\frac{9}{8}}\right)+\mathrm{C}\\ =\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{8\mathrm{x}-9}{9}\right)+\mathrm{C}\\ \mathrm{Hence},\text{\hspace{0.17em}the correct option is B.}\end{array}$

Q.37

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\left(\mathrm{A}+\mathrm{B}\right)\mathrm{x}+2\mathrm{A}+\mathrm{B}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{B}=1\text{and}2\mathrm{A}+\mathrm{B}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=-1\text{and B}=2\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{-1}{\mathrm{x}+1}+\frac{2}{\mathrm{x}+2}\\ \therefore \int \frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}\mathrm{dx}=\int \frac{-1}{\mathrm{x}+1}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{2}{\mathrm{x}+2}\text{\hspace{0.17em}}\mathrm{dx}\\ =\mathrm{log}{\left(\mathrm{x}+1\right)}^{-1}+\mathrm{log}{\left(\mathrm{x}+2\right)}^{2}+\mathrm{C}\\ =\mathrm{log}\frac{{\left(\mathrm{x}+2\right)}^{2}}{\left(\mathrm{x}+1\right)}+\mathrm{C}\end{array}$

Q.38

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{2}-9}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{2}-9}=\frac{1}{\left(\mathrm{x}+3\right)\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\mathrm{x}+3}+\frac{\mathrm{B}}{\mathrm{x}-3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}+3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\left(\mathrm{A}+\mathrm{B}\right)\mathrm{x}+3\left(-\mathrm{A}+\mathrm{B}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{B}=0\text{and}3\left(-\mathrm{A}+\mathrm{B}\right)=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=-\frac{1}{6}\text{and B}=\frac{1}{6}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{\left(\mathrm{x}+3\right)\left(\mathrm{x}-3\right)}=\frac{-1}{6\left(\mathrm{x}+3\right)}+\frac{1}{6\left(\mathrm{x}-3\right)}\\ \therefore \int \frac{1}{\left(\mathrm{x}+3\right)\left(\mathrm{x}-3\right)}\mathrm{dx}=\int \frac{-1}{6\left(\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{1}{6\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{6}\mathrm{log}|\mathrm{x}+3|+\frac{1}{6}\mathrm{log}|\mathrm{x}-3|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}\mathrm{log}|\frac{\mathrm{x}-3}{\mathrm{x}+3}|+\mathrm{C}\end{array}$

Q.39

$\text{Integrate the rational function\hspace{0.17em}}\frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\\ \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\\ \text{ }=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-1=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(\mathrm{x}-3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1,2\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}3\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=1,\text{B}=-5\text{and C}=\text{4}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}=\frac{1}{\mathrm{x}-1}+\frac{-5}{\mathrm{x}-2}+\frac{4}{\left(\mathrm{x}-3\right)}\\ \therefore \int \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{5}{\mathrm{x}-2}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{4}{\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}\left|\mathrm{x}-1\right|-5\mathrm{log}\left|\mathrm{x}-2\right|+4\mathrm{log}\left|\mathrm{x}-3\right|+\mathrm{C}\end{array}$

Q.40

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(\mathrm{x}-3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1,2\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}3\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\frac{1}{2},\text{B}=-2\text{and C}=\frac{3}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}=\frac{1}{2\left(\mathrm{x}-1\right)}+\frac{-2}{\mathrm{x}-2}+\frac{3}{2\left(\mathrm{x}-3\right)}\\ \therefore \int \frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}-2\int \frac{1}{\mathrm{x}-2}\text{\hspace{0.17em}}\mathrm{dx}+\frac{3}{2}\int \frac{1}{\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\mathrm{x}-1|-2\mathrm{log}|\mathrm{x}-2|+\frac{3}{2}\mathrm{log}|\mathrm{x}-3|+\mathrm{C}\end{array}$

Q.41

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}}{{\mathrm{x}}^{2}+3\mathrm{x}+2}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}}{{\mathrm{x}}^{2}+3\mathrm{x}+2}=\frac{2\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=\mathrm{A}\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+1\right)\text{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}-2\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=-2,\text{B}=4\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{-2}{\mathrm{x}+1}+\frac{4}{\mathrm{x}+2}\\ \therefore \int \frac{2\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{-2}{\mathrm{x}+1}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{4}{\mathrm{x}+2}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=-2\mathrm{log}|\mathrm{x}+1|+4\mathrm{log}|\mathrm{x}+2|+\mathrm{C}\end{array}$

Q.42

$\text{Integrate the rational function\hspace{0.17em}}\frac{1-{\mathrm{x}}^{2}}{\mathrm{x}\left(1-2\mathrm{x}\right)}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em} }\frac{1-{\mathrm{x}}^{2}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{1}{2}+\frac{1}{2}\left\{\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}\right\}\text{ }...\left(\mathrm{i}\right)\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{1-2\mathrm{x}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2-\mathrm{x}=\mathrm{A}\left(1-2\mathrm{x}\right)+\mathrm{Bx}\\ \mathrm{Substituting}\text{x}=0\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\frac{1}{2}\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=2,\text{B}=3\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{2}{\mathrm{x}}+\frac{3}{1-2\mathrm{x}}\\ \mathrm{From}\text{​ equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{we}\text{​​ have}\\ \int \frac{1-{\mathrm{x}}^{2}}{\mathrm{x}\left(1-2\mathrm{x}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{2}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{2}{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{3}{1-2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{x}+\frac{1}{2}\left(2\mathrm{log}\left|\mathrm{x}\right|-\frac{3}{4}\mathrm{log}\left|1-2\mathrm{x}\right|\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{x}+\mathrm{log}\left|\mathrm{x}\right|-\frac{3}{4}\mathrm{log}\left|1-2\mathrm{x}\right|+\mathrm{C}\end{array}$

Q.43

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}=\frac{\mathrm{Ax}+\mathrm{B}}{{\mathrm{x}}^{2}+1}+\frac{\mathrm{C}}{\mathrm{x}-1}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{x}=\left(\mathrm{Ax}+\mathrm{B}\right)\left(\mathrm{x}-1\right)+\mathrm{C}\left({\mathrm{x}}^{2}+1\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{Ax}}^{2}-\mathrm{Ax}+\mathrm{Bx}-\mathrm{B}+{\mathrm{Cx}}^{2}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{A}+\mathrm{C}\right){\mathrm{x}}^{2}+\left(-\mathrm{A}+\mathrm{B}\right)\mathrm{x}+\left(-\mathrm{B}+\mathrm{C}\right)\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{, x and constant term from both sides,}\\ \text{we get}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{C}=0,\text{}-\text{A}+\text{B}=1\text{and\hspace{0.17em}\hspace{0.17em}}-\mathrm{B}+\mathrm{C}=0\\ \mathrm{On}\text{​ solving these equations, we get}\\ \text{A}=-\frac{1}{2}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}B}=\frac{1}{2}\text{,\hspace{0.17em}\hspace{0.17em}C}=\frac{1}{2}\\ \mathrm{So},\text{from equation}\left(\mathrm{i}\right)\text{, we get}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}=\frac{\frac{1}{2}\left(-\mathrm{x}+1\right)}{{\mathrm{x}}^{2}+1}+\frac{\frac{1}{2}}{\mathrm{x}-1}\\ \therefore \int \frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{\left(-\mathrm{x}\right)}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{2}+1|+\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\frac{1}{2}\mathrm{log}|\mathrm{x}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\mathrm{x}-1|-\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{2}+1|+\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.44

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{{\left(\mathrm{x}-1\right)}^{2}}+\frac{\mathrm{C}}{\left(\mathrm{x}+2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}-1\right)\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+2\right)+\mathrm{C}{\left(\mathrm{x}-1\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} B}=\frac{1}{3}\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{and constant term, we get}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}A}+\text{C}=\text{0}\\ -\text{2A}+2\mathrm{B}+\mathrm{C}=0\\ \mathrm{On}\text{solving, we get}\\ \text{A}=\frac{2}{9}\text{and C}=\frac{-2}{9}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}=\frac{2}{9\left(\mathrm{x}-1\right)}+\frac{1}{3{\left(\mathrm{x}-1\right)}^{2}}-\frac{2}{9\left(\mathrm{x}+2\right)}\\ \therefore \int \frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{2}{9}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{3}\int \frac{1}{{\left(\mathrm{x}-1\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}-\frac{2}{9}\int \frac{1}{\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{2}{9}\mathrm{log}|\mathrm{x}-1|+\frac{1}{3}\frac{-1}{\left(\mathrm{x}-1\right)}-\frac{2}{9}\mathrm{log}|\mathrm{x}+2|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{9}\mathrm{log}|\frac{\mathrm{x}-1}{\mathrm{x}+2}|-\frac{1}{3}\frac{1}{\left(\mathrm{x}-1\right)}+\mathrm{C}\end{array}$

Q.45

$\text{Integrate the rational function\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\mathrm{x}}^{3}-{\mathrm{x}}^{2}-\mathrm{x}+1}$

Ans.

$\begin{array}{l}\mathrm{Given},\text{\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\mathrm{x}}^{3}-{\mathrm{x}}^{2}-\mathrm{x}+1}=\frac{3\mathrm{x}+5}{{\mathrm{x}}^{2}\left(\mathrm{x}-1\right)-1\left(\mathrm{x}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{3\mathrm{x}+5}{\left(\mathrm{x}-1\right)\left({\mathrm{x}}^{2}-1\right)}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\mathrm{x}+5}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}\\ \mathrm{Let},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{{\left(\mathrm{x}-1\right)}^{2}}+\frac{\mathrm{C}}{\left(\mathrm{x}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+5=\mathrm{A}\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)+\mathrm{B}\left(\mathrm{x}+1\right)+\mathrm{C}{\left(\mathrm{x}-1\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{A}\left({\mathrm{x}}^{2}-1\right)+\mathrm{B}\left(\mathrm{x}+1\right)+\mathrm{C}\left({\mathrm{x}}^{2}-2\mathrm{x}+1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=\text{1 in equation}\left(\mathrm{i}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}B}=\frac{8}{2}=4\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{, x and constant term,we get}\\ \text{A}+\text{C}=\text{0 and B}-2\mathrm{C}=3\\ \mathrm{On}\text{solving these equations, we get}\\ \text{A}=-\frac{1}{2}\text{and C}=\frac{1}{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}=\frac{-1}{2\left(\mathrm{x}-1\right)}+\frac{4}{{\left(\mathrm{x}-1\right)}^{2}}+\frac{1}{2\left(\mathrm{x}+1\right)}\\ ⇒\int \frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}=-\frac{1}{2}\int \frac{1}{\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}+4\int \frac{1}{{\left(\mathrm{x}-1\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\mathrm{log}|\mathrm{x}-1|+4\left(\frac{-1}{\mathrm{x}-1}\right)+\frac{1}{2}\mathrm{log}|\mathrm{x}+1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\frac{\mathrm{x}+1}{\mathrm{x}-1}|-\frac{4}{\mathrm{x}-1}+\mathrm{C}\end{array}$

Q.46

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}$

Ans.

$\begin{array}{l}\frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}=\frac{2\mathrm{x}-3}{\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\left(2\mathrm{x}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\left(\mathrm{x}+1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{C}}{\left(2\mathrm{x}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}-3=\mathrm{A}\left(\mathrm{x}-1\right)\left(2\mathrm{x}+3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(2\mathrm{x}+3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1,1\text{and}-\frac{3}{2}\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{A}=\frac{5}{2}\text{, \hspace{0.17em}B}=-\frac{1}{10}\text{and C}=-\frac{24}{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}-3}{\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\left(2\mathrm{x}+3\right)}=\frac{5}{2\left(\mathrm{x}+1\right)}-\frac{1}{10\left(\mathrm{x}-1\right)}-\frac{24}{5\left(2\mathrm{x}+3\right)}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\int \frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{5}{2}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}-\frac{1}{10}\int \frac{1}{\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{24}{5}\int \frac{1}{\left(2\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{2}\mathrm{log}|\left(\mathrm{x}+1\right)|-\frac{1}{10}\mathrm{log}|\left(\mathrm{x}-1\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}-\frac{24}{5×2}\mathrm{log}|\left(2\mathrm{x}+3\right)|+\mathrm{D}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{2}\mathrm{log}|\left(\mathrm{x}+1\right)|-\frac{1}{10}\mathrm{log}|\left(\mathrm{x}-1\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{12}{5}\mathrm{log}|\left(2\mathrm{x}+3\right)|+\mathrm{D}\end{array}$

Q.47

$\text{Integrate the rational function\hspace{0.17em}}\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^{2}-4\right)}$

Ans.

$\begin{array}{l}\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^{2}-4\right)}=\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\\ \text{ \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\left(\mathrm{x}+1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}+2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{x}=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)+\mathrm{C}\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)\text{ }...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1,-2\text{and\hspace{0.17em}\hspace{0.17em}}2\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{A}=\frac{5}{3}\text{,\hspace{0.17em}\hspace{0.17em}B}=-\frac{5}{2}\text{and C}=\frac{5}{6}\\ ⇒\text{\hspace{0.17em}}\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}=\frac{5}{3\left(\mathrm{x}+1\right)}-\frac{5}{2\left(\mathrm{x}+2\right)}+\frac{5}{6\left(\mathrm{x}-2\right)}\\ \therefore \int \frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{5}{3}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}-\frac{5}{2}\int \frac{1}{\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{ }+\frac{5}{6}\int \frac{1}{\left(\mathrm{x}-2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{ \hspace{0.17em}\hspace{0.17em}}=\frac{5}{3}\mathrm{log}\left|\left(\mathrm{x}+1\right)\right|-\frac{5}{2}\mathrm{log}\left|\left(\mathrm{x}+2\right)\right|\\ \text{ }+\frac{5}{6}\mathrm{log}\left|\left(\mathrm{x}-2\right)\right|+\mathrm{D}\end{array}$

Q.48

$\text{Integrate the rational function\hspace{0.17em}}\frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}$

Ans.

$\begin{array}{l}\because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}=\mathrm{x}+\frac{2\mathrm{x}+1}{{\mathrm{x}}^{2}-1}\\ \mathrm{Let},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}+1}{{\mathrm{x}}^{2}-1}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}-1}\\ 2\mathrm{x}+1=\mathrm{A}\left(\mathrm{x}-1\right)+\mathrm{B}\left(\mathrm{x}+1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1\text{and 1 respectively in equation\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right),\text{we get}\\ \text{2}\left(-1\right)+1=\mathrm{A}\left(-1-1\right)+\mathrm{B}\left(-1+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-1=\mathrm{A}\left(-2\right)⇒\mathrm{A}=\frac{1}{2}\\ \mathrm{and}\text{B}=\frac{3}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}=\mathrm{x}+\frac{1}{2\left(\mathrm{x}+1\right)}+\frac{3}{2\left(\mathrm{x}-1\right)}\\ \int \frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}\text{\hspace{0.17em}}\mathrm{dx}=\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}+\frac{3}{2}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}+\frac{1}{2}\mathrm{log}|\mathrm{x}+1|+\frac{3}{2}\mathrm{log}|\mathrm{x}-1|+\mathrm{C}\end{array}$

Q.49

$\text{Integrate the rational function\hspace{0.17em}}\frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}=\frac{\mathrm{A}}{1-\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{1+{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2=\mathrm{A}\left(1+{\mathrm{x}}^{2}\right)+\left(\mathrm{Bx}+\mathrm{C}\right)\left(1-\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2={\mathrm{x}}^{2}\left(\mathrm{A}-\mathrm{B}\right)+\mathrm{x}\left(\mathrm{B}-\mathrm{C}\right)+\left(\mathrm{A}+\mathrm{C}\right)\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}-\mathrm{B}=0\\ \mathrm{B}-\mathrm{C}=0\\ \mathrm{A}+\mathrm{C}=2\\ \text{On solving these equations, we get}\\ \mathrm{A}=1,\text{}\mathrm{B}=1\text{}\mathrm{and}\text{}\mathrm{C}=1\\ \therefore \frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}=\frac{1}{1-\mathrm{x}}+\frac{\mathrm{x}+1}{1+{\mathrm{x}}^{2}}\\ ⇒\int \frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{1-\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{\mathrm{x}+1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{1-\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{\mathrm{x}}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|1-\mathrm{x}|+\frac{1}{2}\mathrm{log}|1+{\mathrm{x}}^{2}|+{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.50

$\text{Integrate the rational function}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}=\frac{\mathrm{A}}{\left(\mathrm{x}+2\right)}+\frac{\mathrm{B}}{{\left(\mathrm{x}+2\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-1=\mathrm{A}\left(\mathrm{x}+2\right)+\mathrm{B}\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}=3\\ 2\mathrm{A}+\mathrm{B}=-1\\ \mathrm{Solving},\text{the above equations, we get}\\ \text{A}=3\text{}\mathrm{and}\text{}\mathrm{B}=-7\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}=\frac{3}{\left(\mathrm{x}+2\right)}-\frac{7}{{\left(\mathrm{x}+2\right)}^{2}}\\ ⇒\int \frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{3}{\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{7}{{\left(\mathrm{x}+2\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{log}|\mathrm{x}+2|-7\left(\frac{-1}{\mathrm{x}+2}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{log}|\mathrm{x}+2|+\frac{7}{\mathrm{x}+2}+\mathrm{C}\end{array}$

Q.51

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{4}-1}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{{\mathrm{x}}^{4}-1}=\frac{1}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-1\right)\left({\mathrm{x}}^{2}+1\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{4}-1}=\frac{\mathrm{A}}{\left(\mathrm{x}+1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\left({\mathrm{x}}^{2}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{x}-1\right)\left({\mathrm{x}}^{2}+1\right)+\mathrm{B}\left(\mathrm{x}+1\right)\left({\mathrm{x}}^{2}+1\right)+\left({\mathrm{x}}^{2}-1\right)\left(\mathrm{Cx}+\mathrm{D}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left({\mathrm{x}}^{3}+\mathrm{x}-{\mathrm{x}}^{2}-1\right)+\mathrm{B}\left({\mathrm{x}}^{3}+\mathrm{x}+{\mathrm{x}}^{2}+1\right)+\mathrm{C}\left({\mathrm{x}}^{3}-\mathrm{x}\right)+\mathrm{D}\left({\mathrm{x}}^{2}-1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}1={\mathrm{x}}^{3}\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)+{\mathrm{x}}^{2}\left(-\mathrm{A}+\mathrm{B}+\mathrm{D}\right)+\mathrm{x}\left(\mathrm{A}+\mathrm{B}-\mathrm{C}\right)+\left(-\mathrm{A}+\mathrm{B}-\mathrm{D}\right)\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{B}+\mathrm{C}=0\\ \text{\hspace{0.17em}}-\mathrm{A}+\mathrm{B}+\mathrm{D}=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{A}+\mathrm{B}-\mathrm{C}=0\\ \text{\hspace{0.17em}}-\mathrm{A}+\mathrm{B}-\mathrm{D}=1\\ \mathrm{Solving},\text{the above equations, we get}\\ \text{A}=-\frac{1}{4},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=\frac{1}{4},\text{\hspace{0.17em}}\mathrm{C}=0\text{and D}=-\frac{1}{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{4}-1}=\frac{-1}{4\left(\mathrm{x}+1\right)}+\frac{1}{4\left(\mathrm{x}-1\right)}+\frac{-1}{2\left({\mathrm{x}}^{2}+1\right)}\\ ⇒\int \frac{1}{{\mathrm{x}}^{4}-1}\text{\hspace{0.17em}}\mathrm{dx}=-\frac{1}{4}\int \frac{1}{\mathrm{x}+1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{4}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|\left(\mathrm{x}+1\right)|+\frac{1}{4}\mathrm{log}|\left(\mathrm{x}-1\right)|-\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\mathrm{log}|\frac{\mathrm{x}-1}{\mathrm{x}+1}|-\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.52

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}=\frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}×\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}-1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{\mathrm{n}}⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1}⇒\frac{\mathrm{dt}}{{\mathrm{nx}}^{\mathrm{n}-1}}=\mathrm{dx}\\ \therefore \int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Let}\text{}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{t}\left(\mathrm{A}+\mathrm{B}\right)-\mathrm{A}\\ \mathrm{Substituting}\text{​ t}=\text{0 and 1 respectively, we get}\\ \mathrm{A}=1\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=-1\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{1}{\mathrm{t}}+\frac{-1}{\mathrm{t}-1}\\ ⇒\int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{\mathrm{n}}\int \left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}-1}\right)\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}-\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}-1}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\mathrm{log}|\mathrm{t}|-\frac{1}{\mathrm{n}}\mathrm{log}|\mathrm{t}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{\mathrm{n}}\mathrm{log}|{\mathrm{x}}^{\mathrm{n}}|-\frac{1}{\mathrm{n}}\mathrm{log}|{\mathrm{x}}^{\mathrm{n}}-1|+\mathrm{C}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\mathrm{log}|\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{x}}^{\mathrm{n}}-1}|+\mathrm{C}\end{array}$

Q.53

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\mathrm{sinx}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}⇒\frac{\mathrm{dt}}{\mathrm{cosx}}=\mathrm{dx}\\ \therefore \int \frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\mathrm{cosx}}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{\mathrm{cosx}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}=\frac{\mathrm{A}}{1-\mathrm{t}}+\frac{\mathrm{B}}{2-\mathrm{t}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}1=\mathrm{A}\left(2-\mathrm{t}\right)+\mathrm{B}\left(1-\mathrm{t}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \mathrm{Substituting}\text{t}=1\text{and 2 respectively in eqution}\left(1\right),\text{\hspace{0.17em}}\mathrm{we}\text{get}\\ \text{A}=\text{1 and B}=-1\\ \therefore \frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}=\frac{1}{1-\mathrm{t}}-\frac{1}{2-\mathrm{t}}\\ \mathrm{So},\\ \int \frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{1-\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}-\int \frac{1}{2-\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|1-\mathrm{t}|+\mathrm{log}|2-\mathrm{t}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{log}|\frac{2-\mathrm{t}}{1-\mathrm{t}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{2-\mathrm{sinx}}{1-\mathrm{sinx}}|+\mathrm{C}\end{array}$

Q.54

$\text{Integrate the rational function\hspace{0.17em}}\frac{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}}\frac{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}=1+\frac{\left(4{\mathrm{x}}^{2}+10\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}\\ \mathrm{Let}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\\ \frac{\left(4{\mathrm{x}}^{2}+10\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}=\frac{\mathrm{Ax}+\mathrm{B}}{\left({\mathrm{x}}^{2}+3\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\left({\mathrm{x}}^{2}+4\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{x}}^{2}+10\text{\hspace{0.17em}}=\left(\mathrm{Ax}+\mathrm{B}\right)\left({\mathrm{x}}^{2}+4\right)+\left(\mathrm{Cx}+\mathrm{D}\right)\left({\mathrm{x}}^{2}+3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{3}\left(\mathrm{A}+\mathrm{C}\right)+{\mathrm{x}}^{2}\left(\mathrm{B}+\mathrm{D}\right)+\mathrm{x}\left(4\mathrm{A}+3\mathrm{C}\right)+\left(4\mathrm{B}+3\mathrm{D}\right)\\ \mathrm{Comparing}{\text{the coefficients of x}}^{\text{3}}{\text{, x}}^{\text{2}}\text{, x and constant terms,}\\ \text{we get\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{A}+\mathrm{C}=0\\ \mathrm{B}+\mathrm{D}=4\\ 4\mathrm{A}+3\mathrm{C}=0\\ 4\mathrm{B}+3\mathrm{D}=10\\ \mathrm{On}\text{​ solving these equations, we get}\\ \text{A}=0,\mathrm{B}=-2,\mathrm{C}=0\text{and}\mathrm{D}=6\\ \therefore \frac{\left(4{\mathrm{x}}^{2}+10\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}=\frac{-2}{\left({\mathrm{x}}^{2}+3\right)}+\frac{6}{\left({\mathrm{x}}^{2}+4\right)}\\ \mathrm{Then},\\ \int \frac{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int 1\text{\hspace{0.17em}}\mathrm{dx}-\left(\int \frac{-2}{\left({\mathrm{x}}^{2}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}+6\int \frac{1}{\left({\mathrm{x}}^{2}+4\right)}\text{\hspace{0.17em}}\mathrm{dx}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{x}+\frac{2}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)-\frac{6}{2}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{x}+\frac{2}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)-3{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\end{array}$

Q.55

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+3\right)}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}, \mathrm{ }\\ \mathrm{ }\frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+3\right)}\\ \mathrm{Let} \mathrm{t}={\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\\ \int \frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+3\right)}\mathrm{ }\mathrm{dx}=\int \frac{2\mathrm{x}}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}\mathrm{ }\frac{\mathrm{dt}}{2\mathrm{x}}\\ \mathrm{ }=\int \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}\mathrm{ }\mathrm{dt}\\ \mathrm{Let} \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}=\frac{\mathrm{A}}{\mathrm{t}+1}+\frac{\mathrm{B}}{\mathrm{t}+3}\\ 1=\mathrm{A}\left(\mathrm{t}+3\right)+\mathrm{B}\left(\mathrm{t}+1\right)\\ \mathrm{Substituting}\mathrm{t}=-1\mathrm{and}-3\mathrm{respectively},\mathrm{we}\mathrm{get}\\ \mathrm{A}=\frac{1}{2} \mathrm{and} \mathrm{B}=-\frac{1}{2}\\ \therefore \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}=\frac{1}{2\left(\mathrm{t}+1\right)}-\frac{1}{2\left(\mathrm{t}+3\right)}\\ ⇒\int \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}\mathrm{ }\mathrm{dx}=\int \frac{1}{2\left(\mathrm{t}+1\right)}\mathrm{ }\mathrm{dt}-\int \frac{1}{2\left(\mathrm{t}+3\right)}\mathrm{ }\mathrm{dx}\\ \mathrm{ }=\frac{1}{2}\mathrm{log}|\mathrm{t}+1|-\frac{1}{2}\mathrm{log}|\mathrm{t}+3|+\mathrm{C}\\ \mathrm{ }=\frac{1}{2}\mathrm{log}|\frac{\mathrm{t}+1}{\mathrm{t}+3}|+\mathrm{C}\\ =\frac{1}{2}\mathrm{log}|\frac{{\mathrm{x}}^{2}+1}{{\mathrm{x}}^{2}+3}|+\mathrm{C}\end{array}$

Q.56

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}=\frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}×\frac{{\mathrm{x}}^{3}}{{\mathrm{x}}^{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{3}}{{\mathrm{x}}^{4}\left({\mathrm{x}}^{4}-1\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{4}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=4{\mathrm{x}}^{4-1}⇒\frac{\mathrm{dt}}{4{\mathrm{x}}^{3}}=\mathrm{dx}\\ \therefore \int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{x}}^{3}}{{\mathrm{x}}^{4}\left({\mathrm{x}}^{4}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\int \frac{{\mathrm{x}}^{3}}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{4{\mathrm{x}}^{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Let}\text{}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}\\ \mathrm{Substituting}\text{​ t}=\text{0 and 1 respectively, we get}\\ \mathrm{A}=-1\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=1\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{-1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\\ ⇒\int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{4}\int \left(-\frac{1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\right)\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=-\frac{1}{4}\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}+\frac{1}{4}\int \frac{1}{\mathrm{t}-1}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|\mathrm{t}|+\frac{1}{4}\mathrm{log}|\mathrm{t}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{4}|+\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{4}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{4}\mathrm{log}|\frac{{\mathrm{x}}^{4}-1}{{\mathrm{x}}^{4}}|+\mathrm{C}\end{array}$

Q.57

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}$

Ans.

$\begin{array}{l}\int \frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{e}}^{\mathrm{x}}⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}}=\frac{\mathrm{dt}}{\mathrm{t}}\\ \therefore \int \frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{\mathrm{t}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\mathrm{dt}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}}{\mathrm{t}\left(\mathrm{t}-1\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Putting}\text{t}=0\text{​ and 1 respectively in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{get}\\ \text{A}=-1\text{and B}=1\\ \therefore \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{-1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\\ ⇒\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\mathrm{dt}=\int \frac{-1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}+\int \frac{1}{\mathrm{t}-1}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|\mathrm{t}|+\mathrm{log}|\mathrm{t}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{\mathrm{t}-1}{\mathrm{t}}|+\mathrm{C}\\ \therefore \int \frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\mathrm{log}|\frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{\mathrm{x}}}|+\mathrm{C}\end{array}$

Q.58

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{\mathrm{xdx}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\text{ }\mathrm{dx}\text{ }\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }\mathrm{log}\left|\frac{{\left(\mathrm{x}-1\right)}^{2}}{\mathrm{x}-2}\right|+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\mathrm{log}\left|\frac{{\left(\mathrm{x}-2\right)}^{2}}{\mathrm{x}-1}\right|+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }\mathrm{log}\left|{\left(\frac{\mathrm{x}-1}{\mathrm{x}-2}\right)}^{2}\right|+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\mathrm{log}\left|\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\right|+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\int \frac{\mathrm{xdx}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{\mathrm{x}-2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}\left(\mathrm{x}-2\right)+\mathrm{B}\left(\mathrm{x}-1\right)}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}-2\right)+\mathrm{B}\left(\mathrm{x}-1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}2,\text{in equation}\left(\mathrm{i}\right),\text{​\hspace{0.17em}we get}\\ \text{A}=-\text{1 and B}=2\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}=\frac{-1}{\mathrm{x}-1}+\frac{2}{\mathrm{x}-2}\\ \therefore \int \frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{-1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{2}{\mathrm{x}-2}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\text{log}|\mathrm{x}-1|+2\mathrm{log}|\mathrm{x}-2|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{{\left(\mathrm{x}-2\right)}^{2}}{\left(\mathrm{x}-1\right)}|+\mathrm{C}\\ \mathrm{Hence},\text{the option}\left(\text{B}\right)\text{is correct.}\end{array}$

Q.59

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{\mathrm{dx}}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }\mathrm{log}\left|\mathrm{x}\right|-\frac{1}{2}\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\mathrm{log}\left|\mathrm{x}\right|+\frac{1}{2}\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }-\mathrm{log}\left|\mathrm{x}\right|+\frac{1}{2}\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\frac{1}{2}\mathrm{log}\left|\mathrm{x}\right|+\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\int \frac{\mathrm{dx}}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\\ \mathrm{Let}\text{}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{\left({\mathrm{x}}^{2}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}\left({\mathrm{x}}^{2}+1\right)+\left(\mathrm{Bx}+\mathrm{C}\right)\mathrm{x}}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left({\mathrm{x}}^{2}+1\right)+\left({\mathrm{Bx}}^{2}+\mathrm{Cx}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1={\mathrm{x}}^{2}\left(\mathrm{A}+\mathrm{B}\right)+\mathrm{Cx}+\mathrm{A}\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}+\mathrm{B}=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{C}=0\text{\hspace{0.17em}and A}=1\\ \mathrm{On}\text{solving these equations, we get}\\ \text{A}=\text{1,\hspace{0.17em} B}=-\text{1 and C}=0\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}=\frac{1}{\mathrm{x}}+\frac{-\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)}\\ \therefore \int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\text{log}|\mathrm{x}|-2\mathrm{log}|{\mathrm{x}}^{2}+1|+\mathrm{C}\\ \mathrm{Thus},\text{the correct option is}\left(\mathrm{A}\right).\end{array}$

Q.60

Integrate the function x sin x.

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \mathrm{xsinx}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{algebraic function i.e. x as a first function and sinx as}\\ \text{second function and integrating by parts, we get}\\ =\mathrm{x}\int \mathrm{sinx}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int \mathrm{sinx}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ =\mathrm{x}\left(-\mathrm{cosx}\right)-\int 1.\left(-\mathrm{cosx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ =-\mathrm{xcosx}+\mathrm{sinx}+\mathrm{C}\end{array}$

Q.61

Integrate the functions
x sin 3x

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \mathrm{xsin}3\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{algebraic function i.e. x as a first function and sin3x as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\int \mathrm{sin}3\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int \mathrm{sin}3\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\left(-\frac{1}{3}\mathrm{cos}3\mathrm{x}\right)-\int 1.\left(-\frac{1}{3}\mathrm{cosx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{3}\mathrm{xcos}3\mathrm{x}+\frac{1}{9}\mathrm{sin}3\mathrm{x}+\mathrm{C}\end{array}$

Q.62

Integrate the functions
x2 ex

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{x}}^{2}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}{\text{algebraic function i.e. x}}^{\text{2}}{\text{as a first function and e}}^{\text{x}}\text{as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}\right)\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}\left({\mathrm{e}}^{\mathrm{x}}\right)-\int 2\mathrm{x}.\left({\mathrm{e}}^{\mathrm{x}}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}{\mathrm{e}}^{\mathrm{x}}-2\left[\mathrm{x}\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\right]+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}{\mathrm{e}}^{\mathrm{x}}-2\left[{\mathrm{xe}}^{\mathrm{x}}-\int 1.{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\right]+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{xe}}^{\mathrm{x}}+2{\mathrm{e}}^{\mathrm{x}}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\left({\mathrm{x}}^{2}-2\mathrm{x}+2\right)+\mathrm{C}\end{array}$

Q.63

Integrate the functions
x logx

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \mathrm{xlogx}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{logx as a first function and}\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ \text{}=\mathrm{logx}\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{}=\mathrm{logx}\left(\frac{{\mathrm{x}}^{2}}{2}\right)-\int \frac{1}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^{2}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}\mathrm{logx}-\int \frac{\mathrm{x}}{2}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}\mathrm{logx}-\frac{{\mathrm{x}}^{2}}{4}+\mathrm{C}\end{array}$

Q.64

Integrate the functions
x log 2x

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \mathrm{xlog}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{log2x as a first function and}\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}2\mathrm{x}\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{log}2\mathrm{x}\right)\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}2\mathrm{x}\left(\frac{{\mathrm{x}}^{2}}{2}\right)-\int \frac{1}{2\mathrm{x}}×2.\left(\frac{{\mathrm{x}}^{2}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}\mathrm{log}2\mathrm{x}-\int \frac{\mathrm{x}}{2}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}\mathrm{log}2\mathrm{x}-\frac{{\mathrm{x}}^{2}}{4}+\mathrm{C}\end{array}$

Q.65

Integrate the functions
x sin-1x

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{xsin}}^{-1}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}={\text{sin}}^{-\text{1}}\text{x}⇒\mathrm{x}=\mathrm{sint}\\ \mathrm{cost}=\sqrt{1-{\mathrm{sin}}^{2}\mathrm{t}}=\sqrt{1-{\mathrm{x}}^{2}}\\ \mathrm{Then},\text{}\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\text{sin}}^{-\text{1}}\text{x}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{\mathrm{cost}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\mathrm{cost}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \mathrm{tsint}\text{\hspace{0.17em}}\mathrm{cost}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\int \mathrm{tsin}2\mathrm{t}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Taking}\text{algebraic function i.e. t as a first function and sin2t as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[\mathrm{t}\int \mathrm{sin}2\mathrm{t}\text{\hspace{0.17em}}\mathrm{dt}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{t}\right)\int \mathrm{sin}2\mathrm{t}\text{\hspace{0.17em}}\mathrm{dt}\right\}\text{\hspace{0.17em}}\mathrm{dt}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[\mathrm{t}\left(-\frac{1}{2}\mathrm{cos}2\mathrm{t}\right)-\int 1.\left(-\frac{1}{2}\mathrm{co}2\mathrm{t}\right)\text{\hspace{0.17em}}\mathrm{dx}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[-\frac{1}{2}\mathrm{tcos}2\mathrm{t}+\frac{1}{4}\mathrm{sin}2\mathrm{t}+\mathrm{C}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{4}\mathrm{t}\left(1-2{\mathrm{sin}}^{2}\mathrm{t}\right)+\frac{1}{8}×2\mathrm{sintcost}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\mathrm{t}\left(2{\mathrm{sin}}^{2}\mathrm{t}-1\right)+\frac{1}{4}\mathrm{sintcost}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\left(2{\mathrm{x}}^{2}-1\right){\mathrm{sin}}^{-1}\mathrm{x}+\frac{1}{4}\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}+\mathrm{C}\end{array}$

Q.66

Integrate the functions
x tan-1x

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{xtan}}^{-1}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{}{\mathrm{tan}}^{-1}\mathrm{x}\text{as a first function and x as second function and}\\ \text{integrating by parts, we get}\\ ={\mathrm{tan}}^{-1}\mathrm{x}\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\right)\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ ={\mathrm{tan}}^{-1}\mathrm{x}\left(\frac{{\mathrm{x}}^{2}}{2}\right)-\int \frac{1}{1+{\mathrm{x}}^{2}}.\left(\frac{{\mathrm{x}}^{2}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ =\frac{{\mathrm{x}}^{2}}{2}{\mathrm{tan}}^{-1}\mathrm{x}-\frac{1}{2}\int \frac{{\mathrm{x}}^{2}}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^{2}}{2}{\mathrm{tan}}^{-1}\mathrm{x}-\frac{1}{2}\int \frac{1+{\mathrm{x}}^{2}-1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^{2}}{2}{\mathrm{tan}}^{-1}\mathrm{x}-\frac{1}{2}\int \frac{1+{\mathrm{x}}^{2}}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^{2}}{2}{\mathrm{tan}}^{-1}\mathrm{x}-\frac{1}{2}\int 1.\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\\ =\frac{{\mathrm{x}}^{2}}{2}{\mathrm{tan}}^{-1}\mathrm{x}-\frac{1}{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.67

Integrate the functions
x cos-1x

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{xcos}}^{-1}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{t}={\mathrm{cos}}^{-\text{1}}\text{x}⇒\mathrm{x}=\mathrm{cost}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{sint}=\sqrt{1-{\mathrm{cos}}^{2}\mathrm{t}}=\sqrt{1-{\mathrm{x}}^{2}}\\ \mathrm{Then},\text{}\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{cos}}^{-\text{1}}\text{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{-1}{\sqrt{1-{\mathrm{x}}^{2}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{\mathrm{sint}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=-\mathrm{sint}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \mathrm{tcost}\left(-\text{\hspace{0.17em}}\mathrm{sint}\right)\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\int \mathrm{tsin}2\mathrm{t}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Taking}\text{algebraic function i.e. t as a first function and sin2t as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\left[\mathrm{t}\int \mathrm{sin}2\mathrm{t}\text{\hspace{0.17em}}\mathrm{dt}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{t}\right)\int \mathrm{sin}2\mathrm{t}\text{\hspace{0.17em}}\mathrm{dt}\right\}\text{\hspace{0.17em}}\mathrm{dt}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\left[\mathrm{t}\left(-\frac{1}{2}\mathrm{cos}2\mathrm{t}\right)-\int 1.\left(-\frac{1}{2}\mathrm{co}2\mathrm{t}\right)\text{\hspace{0.17em}}\mathrm{dx}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\left[-\frac{1}{2}\mathrm{tcos}2\mathrm{t}+\frac{1}{4}\mathrm{sin}2\mathrm{t}+\mathrm{C}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\mathrm{t}\left(1-2{\mathrm{sin}}^{2}\mathrm{t}\right)-\frac{1}{8}×2\mathrm{sintcost}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\mathrm{t}\left(1-2{\mathrm{sin}}^{2}\mathrm{t}\right)-\frac{1}{4}\mathrm{sintcost}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\left(1-2{\mathrm{x}}^{2}\right){\mathrm{sin}}^{-1}\mathrm{x}-\frac{1}{4}\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}+\mathrm{C}\end{array}$

Q.68

Integrate the functions
(sin–1x)2

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int {\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}.1\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{\hspace{0.17em}}{\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\text{as first function and 1 as second function and}\\ \text{integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\int 1\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\frac{\mathrm{d}}{\mathrm{dx}}{\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\int 1\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}-\int \left\{2{\mathrm{sin}}^{-1}\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sin}}^{-1}\mathrm{x}\right).\mathrm{x}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}-\int \frac{2{\mathrm{xsin}}^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}+\int {\mathrm{sin}}^{-1}\mathrm{x}\frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}+{\mathrm{sin}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}-\left(\int \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}+{\mathrm{sin}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{-2\mathrm{x}}-\left\{\int \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{-2\mathrm{x}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{t}=1-{\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=-2\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{-2\mathrm{x}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}+{\mathrm{sin}}^{-1}\mathrm{x}.2\sqrt{\mathrm{t}}-\left\{\int \frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}2\sqrt{\mathrm{t}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}+2{\mathrm{sin}}^{-1}\mathrm{x}.\sqrt{1-{\mathrm{x}}^{2}}-\int \frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}2\sqrt{1-{\mathrm{x}}^{2}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}\mathrm{x}+2{\mathrm{sin}}^{-1}\mathrm{x}.\sqrt{1-{\mathrm{x}}^{2}}-2\int 1\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}{\left({\mathrm{sin}}^{-1}\mathrm{x}\right)}^{2}+2\sqrt{1-{\mathrm{x}}^{2}}{\mathrm{sin}}^{-1}\mathrm{x}-2\mathrm{x}+\mathrm{C}\end{array}$

Q.69

Integrate the functions
x2 logx

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{x}}^{2}\mathrm{logx}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{logx as a first function and}{\mathrm{x}}^{2}\text{as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{logx}\int {\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int {\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{logx}\left(\frac{{\mathrm{x}}^{3}}{3}\right)-\int \frac{1}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^{3}}{3}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{3}}{3}\mathrm{logx}-\int \frac{{\mathrm{x}}^{2}}{3}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{3}}{3}\mathrm{logx}-\frac{{\mathrm{x}}^{3}}{9}+\mathrm{C}\end{array}$

Q.70

$\begin{array}{l}\text{Integrate the functions}\\ \frac{{\mathrm{xcos}}^{–1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \frac{{\mathrm{xcos}}^{–1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{-1}{2}\int {\mathrm{cos}}^{-1}\mathrm{x}\frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\mathrm{x}\text{as first function and}\frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{as second function and}\\ \text{integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\frac{-1}{2}\left\{{\mathrm{cos}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}-\left(\int \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{cos}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}\right)\mathrm{dx}\right\}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{-1}{2}\left\{{\mathrm{cos}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{-2\mathrm{x}}-\left(\int \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{cos}}^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{-2\mathrm{x}}\right)\text{\hspace{0.17em}}\mathrm{dx}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\mathrm{Let}\text{t}=1-{\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=-2\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{-2\mathrm{x}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{2}\left[{\mathrm{cos}}^{-1}\mathrm{x}.2\sqrt{\mathrm{t}}-\left(\int \frac{-1}{\sqrt{1-{\mathrm{x}}^{2}}}2\sqrt{\mathrm{t}}\right)\text{\hspace{0.17em}}\mathrm{dx}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{2}\left[{\mathrm{cos}}^{-1}\mathrm{x}.2\sqrt{1-{\mathrm{x}}^{2}}-\left(\int \frac{-1}{\sqrt{1-{\mathrm{x}}^{2}}}2\sqrt{1-{\mathrm{x}}^{2}}\right)\text{\hspace{0.17em}}\mathrm{dx}\right]\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{2}\left(2{\mathrm{cos}}^{-1}\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}+2\int 1\text{\hspace{0.17em}}\mathrm{dx}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{2}\left(2{\mathrm{cos}}^{-1}\mathrm{x}\sqrt{1-{\mathrm{x}}^{2}}+2\mathrm{x}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\left(\sqrt{1-{\mathrm{x}}^{2}}{\mathrm{cos}}^{-1}\mathrm{x}+\mathrm{x}\right)+\mathrm{C}\end{array}$

Q.71

Integrate the functions x sec2x

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{xsec}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{algebraic function i.e. x as a first function and}{\mathrm{sec}}^{2}\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\int {\mathrm{sec}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int {\mathrm{sec}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\left(\mathrm{tanx}\right)-\int 1.\mathrm{tanx}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{xtanx}-\int \mathrm{tanx}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{xtanx}+\mathrm{log}|\mathrm{cosx}|+\mathrm{C}\end{array}$

Q.72

Integrate the functions tan-1x

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{tan}}^{-1}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{tan}}^{-1}\mathrm{x}\text{\hspace{0.17em}}.1\mathrm{dx}\\ \mathrm{Taking}\text{}{\mathrm{tan}}^{-1}\mathrm{x}\text{as a first function and 1 as second function and}\\ \text{integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\mathrm{tan}}^{-1}\mathrm{x}\int 1\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\right)\int 1\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}={\mathrm{tan}}^{-1}\mathrm{x}\left(\mathrm{x}\right)-\int \frac{1}{1+{\mathrm{x}}^{2}}.\left(\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{x}\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\mathrm{x}-\int \frac{\mathrm{x}}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}=\mathrm{x}\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\mathrm{x}-\frac{1}{2}\mathrm{log}|1+{\mathrm{x}}^{2}|+\mathrm{C}\end{array}$

Q.73

Integrate the functions x (logx)2

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \mathrm{x}{\left(\mathrm{logx}\right)}^{2}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{}{\left(\text{logx}\right)}^{\text{2}}\text{as a first function and}\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\text{logx}\right)}^{\text{2}}\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\left(\text{logx}\right)}^{\text{2}}\right)\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\text{logx}\right)}^{\text{2}}\left(\frac{{\mathrm{x}}^{2}}{2}\right)-\int \frac{2\mathrm{logx}}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^{2}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}{\left(\text{logx}\right)}^{\text{2}}-\int \mathrm{xlogx}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}{\left(\text{logx}\right)}^{\text{2}}-\left[\mathrm{logx}\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}{\left(\text{logx}\right)}^{\text{2}}-\left\{\mathrm{logx}\left(\frac{{\mathrm{x}}^{2}}{2}\right)-\int \frac{1}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^{2}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}{\left(\text{logx}\right)}^{\text{2}}-\left\{\frac{{\mathrm{x}}^{2}}{2}\mathrm{logx}-\int \frac{\mathrm{x}}{2}\text{\hspace{0.17em}}\mathrm{dx}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}{\left(\text{logx}\right)}^{\text{2}}-\frac{{\mathrm{x}}^{2}}{2}\mathrm{logx}+\frac{{\mathrm{x}}^{2}}{4}+\mathrm{C}\end{array}$

Q.74

Integrate the functions (x2+1) logx

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \left({\mathrm{x}}^{2}+1\right)\mathrm{logx}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}\text{logx as a first function and}\left({\mathrm{x}}^{2}\text{+1}\right)\text{as second function}\\ \text{and integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{logx}\int \left({\mathrm{x}}^{2}+1\right)\text{\hspace{0.17em}}\mathrm{dx}-\int \left\{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int \left({\mathrm{x}}^{2}+1\right)\text{\hspace{0.17em}}\mathrm{dx}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{logx}\left(\frac{{\mathrm{x}}^{3}}{3}+\mathrm{x}\right)-\int \frac{1}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^{3}}{3}+\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{{\mathrm{x}}^{3}}{3}+\mathrm{x}\right)\mathrm{logx}-\int \left(\frac{{\mathrm{x}}^{2}}{3}+1\right)\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{{\mathrm{x}}^{3}}{3}+\mathrm{x}\right)\mathrm{logx}-\frac{{\mathrm{x}}^{3}}{9}-\mathrm{x}+\mathrm{C}\end{array}$

Q.75

Integrate the functions ex (sinx + cosx)

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\left(\mathrm{sinx}+\mathrm{cosx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Here},\text{f}\left(\mathrm{x}\right)=\mathrm{sinx}\text{and f’}\left(\mathrm{x}\right)=\mathrm{cosx}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}\left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right].\\ \mathrm{Therefore},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\left(\mathrm{sinx}+\mathrm{cosx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int {\mathrm{e}}^{\mathrm{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right\}\text{\hspace{0.17em}}\mathrm{dx}={\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{C}\right]\end{array}$

Q.76

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{{\mathrm{xe}}^{\mathrm{x}}}{{\left(1+\mathrm{x}\right)}^{2}}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \frac{{\mathrm{xe}}^{\mathrm{x}}}{{\left(1+\mathrm{x}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{\mathrm{x}}{{\left(1+\mathrm{x}\right)}^{2}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{1+\mathrm{x}-1}{{\left(1+\mathrm{x}\right)}^{2}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{1+\mathrm{x}}{{\left(1+\mathrm{x}\right)}^{2}}-\frac{1}{{\left(1+\mathrm{x}\right)}^{2}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{1}{\left(1+\mathrm{x}\right)}-\frac{1}{{\left(1+\mathrm{x}\right)}^{2}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{​ f}\left(\mathrm{x}\right)=\frac{1}{\left(1+\mathrm{x}\right)}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=-\frac{1}{{\left(1+\mathrm{x}\right)}^{2}}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}\left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right].\\ \mathrm{Therefore},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \frac{{\mathrm{xe}}^{\mathrm{x}}}{{\left(1+\mathrm{x}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{1}{\left(1+\mathrm{x}\right)}-\frac{1}{{\left(1+\mathrm{x}\right)}^{2}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{{\mathrm{e}}^{\mathrm{x}}}{1+\mathrm{x}}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because {\text{e}}^{\text{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\right]\end{array}$

Q.77

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}{\mathrm{e}}^{\mathrm{x}}\left(\frac{1+\mathrm{sinx}}{1+\mathrm{cosx}}\right)$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ {\mathrm{e}}^{\mathrm{x}}\left(\frac{1+\mathrm{sinx}}{1+\mathrm{cosx}}\right)={\mathrm{e}}^{\mathrm{x}}\left(\frac{{\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}+{\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}+2\mathrm{sin}\frac{\mathrm{x}}{2}\mathrm{cos}\frac{\mathrm{x}}{2}}{2{\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}{\left(\frac{\mathrm{sin}\frac{\mathrm{x}}{2}+\mathrm{cos}\frac{\mathrm{x}}{2}}{\mathrm{cos}\frac{\mathrm{x}}{2}}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}{\left(\mathrm{tan}\frac{\mathrm{x}}{2}+1\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}\left(1+{\mathrm{tan}}^{2}\frac{\mathrm{x}}{2}+2\mathrm{tan}\frac{\mathrm{x}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}\left({\mathrm{sec}}^{2}\frac{\mathrm{x}}{2}+2\mathrm{tan}\frac{\mathrm{x}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{2}{\mathrm{sec}}^{2}\frac{\mathrm{x}}{2}+\mathrm{tan}\frac{\mathrm{x}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\left(\mathrm{tan}\frac{\mathrm{x}}{2}+\frac{1}{2}{\mathrm{sec}}^{2}\frac{\mathrm{x}}{2}\right)\\ \mathrm{Let}\text{f}\left(\mathrm{x}\right)=\mathrm{tan}\frac{\mathrm{x}}{2}\text{\hspace{0.17em} and f’}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{sec}}^{2}\frac{\mathrm{x}}{2}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}\left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right].\\ \mathrm{Therefore},\\ \int {\mathrm{e}}^{\mathrm{x}}\left(\frac{1+\mathrm{sinx}}{1+\mathrm{cosx}}\right)\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{x}}\left(\mathrm{tan}\frac{\mathrm{x}}{2}+\frac{1}{2}{\mathrm{sec}}^{2}\frac{\mathrm{x}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\mathrm{tan}\frac{\mathrm{x}}{2}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because {\text{e}}^{\text{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\right]\end{array}$

Q.78

$\mathrm{Integrate}\mathrm{the}\mathrm{functionse}{}^{\mathrm{x}}\left(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^{2}}\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^{2}}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{​ f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=-\frac{1}{{\mathrm{x}}^{2}}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}\left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right].\\ \mathrm{Therefore},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^{2}}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because {\text{e}}^{\text{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\right]\end{array}$

Q.79

$\begin{array}{l}\text{Integrate the functions}\\ \frac{\left(\mathrm{x}-3\right){\mathrm{e}}^{\mathrm{x}}}{{\left(\mathrm{x}-1\right)}^{3}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \frac{\left(\mathrm{x}-3\right){\mathrm{e}}^{\mathrm{x}}}{{\left(\mathrm{x}-1\right)}^{3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{\mathrm{x}-1-2}{{\left(1-\mathrm{x}\right)}^{3}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{\mathrm{x}-1}{{\left(1-\mathrm{x}\right)}^{3}}-\frac{2}{{\left(1-\mathrm{x}\right)}^{3}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{1}{{\left(1-\mathrm{x}\right)}^{2}}-\frac{2}{{\left(1-\mathrm{x}\right)}^{3}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{​ f}\left(\mathrm{x}\right)=\frac{1}{{\left(1-\mathrm{x}\right)}^{2}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=-\frac{2}{{\left(1-\mathrm{x}\right)}^{3}}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}\left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right].\\ \mathrm{Therefore},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \frac{\left(\mathrm{x}-3\right){\mathrm{e}}^{\mathrm{x}}}{{\left(\mathrm{x}-1\right)}^{3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{1}{{\left(1-\mathrm{x}\right)}^{2}}-\frac{2}{{\left(1-\mathrm{x}\right)}^{3}}\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{e}}^{\mathrm{x}}}{{\left(1-\mathrm{x}\right)}^{2}}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because {\text{e}}^{\text{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\right]\end{array}$

Q.80

Integrate the functions e2x sinx

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Taking}{\text{sinx as first function and e}}^{\text{2x}}\text{as second function, then}\\ \mathrm{Integrating}\text{by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{sinx}\int {\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\int \left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\int {\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{sinx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)-\int \left\{\mathrm{cosx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)\right\}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\int \mathrm{cosx}.{\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\left\{\mathrm{cosx}\int {\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\int \left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\int {\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\right)\text{\hspace{0.17em}}\mathrm{dx}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\left\{\mathrm{cosx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)-\int \left\{-\mathrm{sinx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)\right\}\text{\hspace{0.17em}}\mathrm{dx}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\mathrm{cosx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)-\frac{1}{4}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\text{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}\text{\hspace{0.17em}}=\frac{{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}}{2}-\frac{1}{4}\mathrm{cosx}\text{\hspace{0.17em}}{\mathrm{e}}^{2\mathrm{x}}-\frac{1}{4}\mathrm{I}+\mathrm{C}\\ \mathrm{I}+\frac{1}{4}\mathrm{I}=\frac{{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}}{2}-\frac{1}{4}\mathrm{cosx}\text{\hspace{0.17em}}{\mathrm{e}}^{2\mathrm{x}}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{5}{4}\mathrm{I}=\frac{{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}}{2}-\frac{1}{4}\mathrm{cosx}\text{\hspace{0.17em}}{\mathrm{e}}^{2\mathrm{x}}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\frac{4}{5}\frac{{\mathrm{e}}^{2\mathrm{x}}}{4}\left(2\mathrm{sinx}-\mathrm{cosx}\text{\hspace{0.17em}}\right)+\mathrm{C}\end{array}$

Q.81

$\begin{array}{l}\text{Integrate the functions}\\ {\mathrm{sin}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\mathrm{sin}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}\right)\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{sin}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{x}=\mathrm{tan\theta }⇒\frac{\mathrm{dx}}{\mathrm{d\theta }}={\mathrm{sec}}^{2}\mathrm{\theta }\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{sin}}^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{sin}}^{–1}\left(\frac{2\mathrm{tan\theta }}{1+{\mathrm{tan}}^{2}\mathrm{\theta }}\right)\text{\hspace{0.17em}}{\mathrm{sec}}^{2}\mathrm{\theta }\text{\hspace{0.17em}}\mathrm{d\theta }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{sin}}^{–1}\left(\mathrm{sin}2\mathrm{\theta }\right){\mathrm{sec}}^{2}\mathrm{\theta }\text{\hspace{0.17em}}\mathrm{d\theta }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int 2\mathrm{\theta }\text{\hspace{0.17em}}{\mathrm{sec}}^{2}\mathrm{\theta }\text{\hspace{0.17em}}\mathrm{d\theta }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}=2\left[\mathrm{\theta }\int {\mathrm{sec}}^{2}\mathrm{\theta }\text{\hspace{0.17em}}\mathrm{d\theta }-\left(\int \frac{\mathrm{d}}{\mathrm{d\theta }}\text{\hspace{0.17em}}\mathrm{\theta }\int {\mathrm{sec}}^{2}\mathrm{\theta }\text{\hspace{0.17em}}\mathrm{d\theta }\right)\text{\hspace{0.17em}}\mathrm{d\theta }\right]\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left[\mathrm{\theta tan\theta }-\int 1\mathrm{tan\theta }\text{\hspace{0.17em}}\mathrm{d\theta }\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}=2\mathrm{\theta tan\theta }-2\mathrm{log}|\mathrm{sec\theta }|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{\theta tan\theta }-2\mathrm{log}|\sqrt{1+{\mathrm{tan}}^{2}\mathrm{\theta }}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=2{\mathrm{xtan}}^{-1}\mathrm{x}-2\mathrm{log}|\sqrt{1+{\mathrm{x}}^{2}}|+\mathrm{C}.\end{array}$

Q.82

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int {\mathrm{x}}^{2}{\mathrm{e}}^{{\mathrm{x}}^{3}}\text{ }\mathrm{dx}\text{ }\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }\frac{1}{3}{\mathrm{e}}^{{\mathrm{x}}^{3}}+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\frac{1}{3}{\mathrm{e}}^{{\mathrm{x}}^{2}}+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^{3}}+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^{2}}+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{have}\text{\hspace{0.17em}\hspace{0.17em}}\int {\mathrm{x}}^{2}{\mathrm{e}}^{{\mathrm{x}}^{3}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{​ t}={\text{x}}^{\text{3}}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^{2}\\ \therefore \int {\mathrm{x}}^{2}{\mathrm{e}}^{{\mathrm{x}}^{3}}\text{\hspace{0.17em}}\mathrm{dx}=\int {\mathrm{x}}^{2}{\mathrm{e}}^{\mathrm{t}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{3{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{3}\int {\mathrm{e}}^{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}{\mathrm{e}}^{{\mathrm{x}}^{3}}+\mathrm{C}\\ \mathrm{Hence},\text{the correct option is A.}\end{array}$

Q.83

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int {\mathrm{e}}^{\mathrm{x}}\text{ }\mathrm{secx}\left(1+\mathrm{tanx}\right)\mathrm{dx}\text{ }\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }{\mathrm{e}}^{\mathrm{x}}\mathrm{secx}+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }{\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }{\mathrm{e}}^{\mathrm{x}}\mathrm{tanx}+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{secx}\left(1+\mathrm{tanx}\right)\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\left(\mathrm{secx}+\mathrm{secxtanx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{​ f}\left(\mathrm{x}\right)=\mathrm{secx}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=\mathrm{secxtanx}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}\left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right].\\ \mathrm{Therefore},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\left(\mathrm{secx}+\mathrm{secxtanx}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\mathrm{secx}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because {\text{e}}^{\text{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\right\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\right]\\ \mathrm{Hence},\text{the correct option is B.}\end{array}$

Q.84

$\int \sqrt{4-{\mathrm{x}}^{2}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{4-{\mathrm{x}}^{2}}\\ \mathrm{Let}\text{I}=\int \sqrt{4-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{2}^{2}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}\right]\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{x}\sqrt{{2}^{2}-{\mathrm{x}}^{2}}+\frac{{2}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{x}}{2}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}}{2}\sqrt{4-{\mathrm{x}}^{2}}+2{\mathrm{sin}}^{-1}\frac{\mathrm{x}}{2}+\mathrm{C}\end{array}$

Q.85

$\int \sqrt{1-4{\mathrm{x}}^{2}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{1-4{\mathrm{x}}^{2}}\\ \mathrm{Let}\text{I}=\int \sqrt{1-4{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{1}^{2}-{\left(2\mathrm{x}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \left[\mathrm{Let}\text{t}=2\mathrm{x}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2⇒\mathrm{dx}=\frac{\mathrm{dt}}{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{1}^{2}-{\mathrm{t}}^{2}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(\frac{1}{2}\mathrm{t}\sqrt{{1}^{2}-{\mathrm{t}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{t}}{1}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(\frac{2\mathrm{x}}{2}\sqrt{1-4{\mathrm{x}}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\frac{2\mathrm{x}}{1}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}}{2}\sqrt{1-4{\mathrm{x}}^{2}}+\frac{1}{4}{\mathrm{sin}}^{-1}2\mathrm{x}+\mathrm{C}\end{array}$

Q.86

$\int \sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+6}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+6}=\sqrt{{\left(\mathrm{x}+2\right)}^{2}+2}\\ \text{Let I}=\int \sqrt{{\left(\mathrm{x}+2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\mathrm{t}}^{2}+{\left(\sqrt{2}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^{2}+{\left(\sqrt{2}\right)}^{2}}+\frac{{\left(\sqrt{2}\right)}^{2}}{2}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}+{\left(\sqrt{2}\right)}^{2}}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+2,\text{\hspace{0.17em}we\hspace{0.17em}get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{{\left(\mathrm{x}+2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}+\frac{2}{2}\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\left(\mathrm{x}+2\right)}^{2}+{\left(\sqrt{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+6}+\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+6}|+\mathrm{C}\end{array}$

Q.87

$\int \sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+1}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{We have,}\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+1}=\sqrt{{\left(\mathrm{x}+2\right)}^{2}-3}\\ \text{Let I}=\int \sqrt{{\left(\mathrm{x}+2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\mathrm{t}}^{2}-{\left(\sqrt{3}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}-\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{{\left(\sqrt{3}\right)}^{2}}{2}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}-{\left(\sqrt{3}\right)}^{2}}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+2,\text{\hspace{0.17em}we\hspace{0.17em}get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{{\left(\mathrm{x}+2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{3}{2}\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\left(\mathrm{x}+2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+1}-\frac{3}{2}\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+1}|+\mathrm{C}\end{array}$

Q.88

$\int \sqrt{1-4\mathrm{x}-{\mathrm{x}}^{2}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{1-4\mathrm{x}-{\mathrm{x}}^{2}}=\sqrt{5-\left({\mathrm{x}}^{2}+4\mathrm{x}+4\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(\sqrt{5}\right)}^{2}-{\left(\mathrm{x}+2\right)}^{2}}\\ \mathrm{Let}\text{I}=\int \sqrt{{\left(\sqrt{5}\right)}^{2}-{\left(\mathrm{x}+2\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\left(\sqrt{5}\right)}^{2}-{\mathrm{t}}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{x}}{\mathrm{a}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\left(\sqrt{5}\right)}^{2}-{\mathrm{t}}^{2}}+\frac{{\left(\sqrt{5}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{t}}{\left(\sqrt{5}\right)}+\mathrm{C}\\ \mathrm{Putting}\text{t}=\mathrm{x}+2,\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{{\left(\sqrt{5}\right)}^{2}-{\left(\mathrm{x}+2\right)}^{2}}+\frac{{\left(\sqrt{5}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\left(\mathrm{x}+2\right)}{\left(\sqrt{5}\right)}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{1-4\mathrm{x}-{\mathrm{x}}^{2}}+\frac{5}{2}{\mathrm{sin}}^{-1}\frac{\left(\mathrm{x}+2\right)}{\left(\sqrt{5}\right)}+\mathrm{C}\end{array}$

Q.89

$\int \sqrt{{\mathrm{x}}^{2}+4\mathrm{x}-5}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{We have,}\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}-5}=\sqrt{{\left(\mathrm{x}+2\right)}^{2}-9}\\ \text{Let I}=\int \sqrt{{\left(\mathrm{x}+2\right)}^{2}-{\left(3\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}-\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}-\frac{{\left(3\right)}^{2}}{2}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+2,\text{\hspace{0.17em}we\hspace{0.17em}get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{{\left(\mathrm{x}+2\right)}^{2}-{\left(3\right)}^{2}}-\frac{9}{2}\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\left(\mathrm{x}+2\right)}^{2}-{\left(3\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+2\right)}{2}\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}-5}-\frac{9}{2}\mathrm{log}|\left(\mathrm{x}+2\right)+\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}-5}|+\mathrm{C}\end{array}$

Q.90

$\int \sqrt{1+3\mathrm{x}-{\mathrm{x}}^{2}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{1+3\mathrm{x}-{\mathrm{x}}^{2}}=\sqrt{1-\left({\mathrm{x}}^{2}-3\mathrm{x}+\frac{9}{4}-\frac{9}{4}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^{2}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}\\ \mathrm{Let}\text{I}=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^{2}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^{2}-{\mathrm{t}}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}-\frac{3}{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{x}}{\mathrm{a}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^{2}-{\mathrm{t}}^{2}}+\frac{{\left(\frac{\sqrt{13}}{2}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\mathrm{t}}{\left(\frac{\sqrt{13}}{2}\right)}+\mathrm{C}\\ \text{Putting t}=\mathrm{x}-\frac{3}{2},\text{\hspace{0.17em}we\hspace{0.17em}get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}-\frac{3}{2}\right)}{2}\sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^{2}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}+\frac{{\left(\frac{\sqrt{13}}{2}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\left(\mathrm{x}-\frac{3}{2}\right)}{\left(\frac{\sqrt{13}}{2}\right)}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(2\mathrm{x}-3\right)}{4}\sqrt{1+3\mathrm{x}-{\mathrm{x}}^{2}}+\frac{13}{8}{\mathrm{sin}}^{-1}\frac{\left(2\mathrm{x}-3\right)}{\left(\sqrt{13}\right)}+\mathrm{C}\end{array}$

Q.91

$\int \sqrt{{\mathrm{x}}^{2}+3\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{We have,}\sqrt{{\mathrm{x}}^{2}+3\mathrm{x}}=\sqrt{\left({\mathrm{x}}^{2}+3\mathrm{x}+\frac{9}{4}\right)-\frac{9}{4}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(\mathrm{x}+\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}}\\ \text{Let I}=\int \sqrt{{\left(\mathrm{x}+\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\mathrm{t}}^{2}-{\left(\frac{3}{2}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Let\hspace{0.17em}t}=\mathrm{x}+\frac{3}{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}-\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^{2}-{\left(\frac{3}{2}\right)}^{2}}-\frac{{\left(\frac{3}{2}\right)}^{2}}{2}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}-{\left(\frac{3}{2}\right)}^{2}}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+\frac{3}{2},\text{\hspace{0.17em}we\hspace{0.17em}get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}+\frac{3}{2}\right)}{2}\sqrt{{\left(\mathrm{x}+\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}}-\frac{9}{8}\mathrm{log}|\left(\mathrm{x}+\frac{3}{2}\right)+\sqrt{{\left(\mathrm{x}+\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(2\mathrm{x}+3\right)}{4}\sqrt{{\mathrm{x}}^{2}+3\mathrm{x}}-\frac{9}{8}\mathrm{log}|\left(\mathrm{x}+\frac{3}{2}\right)+\sqrt{{\mathrm{x}}^{2}+3\mathrm{x}}|+\mathrm{C}\end{array}$

Q.92

$\int \sqrt{1+\frac{{\mathrm{x}}^{2}}{9}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}}\mathrm{have},\text{\hspace{0.17em}}\sqrt{1+\frac{{\mathrm{x}}^{2}}{9}}=\frac{1}{3}\sqrt{9+{\mathrm{x}}^{2}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\int \frac{1}{3}\sqrt{9+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{3}\int \sqrt{{3}^{2}+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{3}\int \sqrt{{\mathrm{x}}^{2}+{3}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\left(\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+{3}^{2}}+\frac{{3}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{3}^{2}}|\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}}{6}\sqrt{{\mathrm{x}}^{2}+9}+\frac{3}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+9}|+\mathrm{C}\end{array}$

Q.93

$\begin{array}{l}\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \sqrt{1+{\text{x}}^{2}}\text{ }\mathrm{dx}\mathrm{is}\mathrm{equal}\mathrm{to}\end{array}\\ \left(\text{A}\right)\text{ }\frac{\text{x}}{2}\sqrt{1+{\text{x}}^{2}}+\frac{1}{2}\mathrm{log}\left|\left(\text{x}+\sqrt{1+{\text{x}}^{2}}\right)\right|+\text{C}\\ \left(\text{B}\right)\text{ }\frac{2}{3}{\left(1+{\text{x}}^{2}\right)}^{\frac{3}{2}}+\text{C}\\ \left(\text{C}\right)\text{ }\frac{2}{3}\text{x}{\left(1+{\text{x}}^{2}\right)}^{\frac{3}{2}}+\text{C}\\ \left(\text{D}\right)\text{ }\frac{{\text{x}}^{2}}{2}\sqrt{1+{\text{x}}^{2}}\text{ }+\frac{1}{2}{\text{ x}}^{2}\text{ }\mathrm{log}\left|\left(\text{x}+\sqrt{1+{\text{x}}^{2}}\right)\right|+\text{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \sqrt{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\mathrm{x}}^{2}+{1}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+{1}^{2}}+\frac{{1}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{1}^{2}}|\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+1}+\frac{1}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}}{2}\sqrt{1+{\mathrm{x}}^{2}}+\frac{1}{2}\mathrm{log}|\mathrm{x}+\sqrt{1+{\mathrm{x}}^{2}}|+\mathrm{C}\\ \mathrm{Hence},\text{​\hspace{0.17em}\hspace{0.17em}}\mathrm{the}\text{correct option is A.}\end{array}$

Q.94

$\begin{array}{l}\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\text{ }\mathrm{dx}\mathrm{is}\mathrm{equal}\mathrm{to}\end{array}\\ \left(\mathrm{A}\right)\text{ }\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}+9\text{ }\mathrm{log}\left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\frac{1}{4}\left(\mathrm{x}+4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}+9\mathrm{log}\left|\mathrm{x}+4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}-3\sqrt{2}\text{ }\mathrm{log}\left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}-\frac{9}{2}\mathrm{log}\left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}=\sqrt{{\left(\mathrm{x}-4\right)}^{2}-9}\\ \mathrm{Let}\text{I}=\int \sqrt{{\left(\mathrm{x}-4\right)}^{2}-{\left(3\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}-4⇒\frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \int \sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}-\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}-\frac{{\left(3\right)}^{2}}{2}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}|+\mathrm{C}\\ \mathrm{Putting}\text{t}=\mathrm{x}-4,\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}-4\right)}{2}\sqrt{{\left(\mathrm{x}-4\right)}^{2}-{\left(3\right)}^{2}}-\frac{9}{2}\mathrm{log}|\left(\mathrm{x}-4\right)+\sqrt{{\left(\mathrm{x}-4\right)}^{2}-{\left(3\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}-4\right)}{2}\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}-\frac{9}{2}\mathrm{log}|\left(\mathrm{x}-4\right)+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}|+\mathrm{C}\\ \mathrm{Hence},\text{the correct option is D.}\end{array}$

Q.95

$\int \mathrm{x}\sqrt{\mathrm{x}+{\mathrm{x}}^{2}}\mathrm{dx}$

Ans.

$\begin{array}{l}\int \mathrm{x}\sqrt{\mathrm{x}+{\mathrm{x}}^{2}}\mathrm{dx}\\ \mathrm{Let}\text{x}=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+{\mathrm{x}}^{2}\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{A}\left(1+2\mathrm{x}\right)+\mathrm{B}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=2\mathrm{A}\text{& A}+\text{B}=\text{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\frac{1}{2}\text{&\hspace{0.17em}B}=-\frac{1}{2}\\ \therefore \int \mathrm{x}\sqrt{\mathrm{x}+{\mathrm{x}}^{2}}\mathrm{dx}=\frac{1}{2}\int \left(1+\mathrm{x}\right)\sqrt{\mathrm{x}+{\mathrm{x}}^{2}}\mathrm{dx}-\frac{1}{2}\int \sqrt{\mathrm{x}+{\mathrm{x}}^{2}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×\frac{{\left(\mathrm{x}+{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-\frac{1}{2}\int \sqrt{{\left(\mathrm{x}+\frac{1}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}×{\left(\mathrm{x}+{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}-\frac{1}{2}×\frac{\left(\mathrm{x}+\frac{1}{2}\right)}{2}\sqrt{{\left(\mathrm{x}+\frac{1}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\frac{1}{2}×\frac{{\left(\frac{1}{2}\right)}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\left(\mathrm{x}+\frac{1}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(\mathrm{x}+{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}{3}-\frac{1}{8}\left(2\mathrm{x}+1\right)\sqrt{\mathrm{x}+{\mathrm{x}}^{2}}+\frac{1}{16}\mathrm{log}|\mathrm{x}+\sqrt{\mathrm{x}+{\mathrm{x}}^{2}}|+\mathrm{C}\end{array}$

Q.96

$\int \left(\mathrm{x}+1\right)\sqrt{2{\mathrm{x}}^{2}+3}\mathrm{dx}$

Ans.

$\begin{array}{l}\int \left(\mathrm{x}+1\right)\sqrt{2{\mathrm{x}}^{2}+3}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{x}+1=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\mathrm{x}}^{2}+3\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}+1\text{\hspace{0.17em}}=\mathrm{A}\left(4\mathrm{x}+0\right)+\mathrm{B}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}+1=4\mathrm{Ax}\text{\hspace{0.17em}\hspace{0.17em} & B}=\text{1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\frac{1}{4}\text{&\hspace{0.17em}B}=1\\ \therefore \int \left(\mathrm{x}+1\right)\sqrt{2{\mathrm{x}}^{2}+3}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{4}\int 4\mathrm{x}\sqrt{2{\mathrm{x}}^{2}+3}\text{\hspace{0.17em}}\mathrm{dx}+1\int \sqrt{2{\mathrm{x}}^{2}+3}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}×\frac{{\left(2{\mathrm{x}}^{2}+3\right)}^{\frac{3}{2}}}{\frac{3}{2}}+\sqrt{2}\text{\hspace{0.17em}}\int \sqrt{{\mathrm{x}}^{2}+{\left(\sqrt{\frac{3}{2}}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}×{\left(2{\mathrm{x}}^{2}+3\right)}^{\frac{3}{2}}+\sqrt{2}\text{\hspace{0.17em}}\int \sqrt{{\mathrm{x}}^{2}+{\left(\sqrt{\frac{3}{2}}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}×{\left(2{\mathrm{x}}^{2}+3\right)}^{\frac{3}{2}}+\sqrt{2}\left(\begin{array}{l}\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{x}}^{2}+{\left(\sqrt{\frac{3}{2}}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{{\left(\sqrt{\frac{3}{2}}\right)}^{2}}{2}\mathrm{log}|\mathrm{x}+{\left(\sqrt{\frac{3}{2}}\right)}^{2}|\end{array}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}×{\left(2{\mathrm{x}}^{2}+3\right)}^{\frac{3}{2}}+\frac{\sqrt{2}}{2}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\sqrt{{\mathrm{x}}^{2}+\frac{3}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\frac{3\sqrt{2}}{4}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{\left(\sqrt{\frac{3}{2}}\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(2{\mathrm{x}}^{2}+3\right)}^{\frac{3}{2}}}{6}-\frac{\mathrm{x}}{2}\text{\hspace{0.17em}}\sqrt{2{\mathrm{x}}^{2}+3}-\frac{3\sqrt{2}}{4}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+\frac{3}{2}}|+\mathrm{C}\end{array}$

Q.97

$\left(\mathrm{x}+3\right)\sqrt{3–4\mathrm{x}–{\mathrm{x}}^{2}}$

Ans.

$\begin{array}{l}\int \left(\mathrm{x}+3\right)\sqrt{3-4\mathrm{x}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{x}+3=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(3-4\mathrm{x}-{\mathrm{x}}^{2}\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}+3\text{\hspace{0.17em}}=\mathrm{A}\left(-\text{\hspace{0.17em}}4-2\mathrm{x}\right)+\mathrm{B}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3=-\text{\hspace{0.17em}}4\mathrm{A}+\mathrm{B}\text{\hspace{0.17em}\hspace{0.17em} &}-2\text{A}=\text{1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\frac{1}{-2}\text{}\\ \text{&\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}B}=3+4\mathrm{A}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3+4\left(\frac{1}{-2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3-2=1\\ \therefore \int \left(\mathrm{x}+3\right)\sqrt{3-4\mathrm{x}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{-2}\int \left(-\text{\hspace{0.17em}}4-2\mathrm{x}\right)\sqrt{3-4\mathrm{x}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+1\int \sqrt{3-4\mathrm{x}-{\mathrm{x}}^{2}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{-2}\frac{{\left(3-4\mathrm{x}-{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}{\frac{3}{2}}+\int \sqrt{7-{\left(\mathrm{x}+2\right)}^{2}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(3-4\mathrm{x}-{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}{-3}+\int \sqrt{{\left(\sqrt{7}\right)}^{2}-{\left(\mathrm{x}+2\right)}^{2}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(3-4\mathrm{x}-{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}{-3}+\frac{1}{2}\left(\mathrm{x}+2\right)\sqrt{{\left(\sqrt{7}\right)}^{2}-{\left(\mathrm{x}+2\right)}^{2}}+\frac{7}{2}{\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{7}}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\frac{{\left(\sqrt{7}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{7}}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(3-4\mathrm{x}-{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}{-3}+\frac{1}{2}\left(\mathrm{x}+2\right)\sqrt{3-4\mathrm{x}-{\mathrm{x}}^{2}}\text{\hspace{0.17em}\hspace{0.17em}}+\frac{7}{2}{\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{7}}\right)+\mathrm{C}\end{array}$

Q.98

$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}.\phantom{\rule{0ex}{0ex}}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{By definition}\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{h}\right)+...+\mathrm{f}\left(\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right],\\ \text{where, h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \text{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \therefore {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{h}\right)+...+\mathrm{f}\left(\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{a}+\mathrm{a}+\mathrm{h}+...+\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{na}+\mathrm{h}+2\mathrm{h}+...+\left(\mathrm{n}-1\right)\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{na}+\left\{1+2+...+\left(\mathrm{n}-1\right)\right\}\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{na}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}}{2}\mathrm{h}\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \sum \mathrm{n}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[\mathrm{a}+\frac{\left(\mathrm{n}-1\right)}{2}\left(\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[\mathrm{a}+\frac{\left(\mathrm{b}-\mathrm{a}\right)}{2}\left(1-\frac{1}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\left[\mathrm{a}+\frac{\left(\mathrm{b}-\mathrm{a}\right)}{2}\left(1-0\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\left(\frac{2\mathrm{a}+\mathrm{b}-\mathrm{a}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{b}-\mathrm{a}\right)\frac{\left(\mathrm{b}+\mathrm{a}\right)}{2}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{b}}^{2}-{\mathrm{a}}^{2}}{2}\end{array}$

Q.99

$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}.\phantom{\rule{0ex}{0ex}}{\int }_{0}^{5}\left(\mathrm{x}+1\right)\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}}\mathrm{have},\text{\hspace{0.17em}\hspace{0.17em}}{\int }_{0}^{5}\left(\mathrm{x}+1\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{By}\text{definition}\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{h}\right)+...+\mathrm{f}\left(\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right],\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=0,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{5-0}{\mathrm{n}}=\frac{5}{\mathrm{n}}\\ \therefore {\int }_{0}^{5}\left(\mathrm{x}+1\right)\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(0\right)+\mathrm{f}\left(0+\mathrm{h}\right)+...+\mathrm{f}\left(0+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\left(0+1\right)+\left(\mathrm{h}+1\right)+...+\left\{\left(\mathrm{n}-1\right)\mathrm{h}+1\right\}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\mathrm{h}+2\mathrm{h}+...+\left(\mathrm{n}-1\right)\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\left\{1+2+...+\left(\mathrm{n}-1\right)\right\}\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}}{2}\mathrm{h}\right]\left[\because \sum \mathrm{n}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[1+\frac{\left(\mathrm{n}-1\right)}{2}\left(\frac{5}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[1+\frac{5}{2}\left(1-\frac{1}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\left[1+\frac{5}{2}\left(1-0\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5×\frac{7}{2}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{35}{2}\end{array}$

Q.100

$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}.\phantom{\rule{0ex}{0ex}}{\int }_{2}^{3}{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{We have,}{\int }_{2}^{3}{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{dx}\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{h}\right)+...+\mathrm{f}\left(\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right],\\ \text{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \text{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=3,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{2},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{3-2}{\mathrm{n}}=\frac{1}{\mathrm{n}}\\ \therefore \text{\hspace{0.17em}}{\int }_{2}^{3}{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3-2\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(2\right)+\mathrm{f}\left(2+\mathrm{h}\right)+...+\mathrm{f}\left(2+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[{\left(2\right)}^{2}+{\left(2+\mathrm{h}\right)}^{2}+...+{\left\{2+\left(\mathrm{n}-1\right)\mathrm{h}\right\}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[4+\left(4+4\mathrm{h}+{\mathrm{h}}^{2}\right)+...+\left\{4+4\left(\mathrm{n}-1\right)\mathrm{h}+{\left(\mathrm{n}-1\right)}^{2}{\mathrm{h}}^{2}\right\}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[4\mathrm{n}+4\left\{1+2+...+\left(\mathrm{n}-1\right)\right\}\mathrm{h}+\left\{1+{2}^{2}+{3}^{2}+...+{\left(\mathrm{n}-1\right)}^{2}\right\}{\mathrm{h}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[4\mathrm{n}+4\frac{\left(\mathrm{n}-1\right)\mathrm{n}}{2}\mathrm{h}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}\left(2\mathrm{n}-2+1\right)}{6}{\mathrm{h}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[4\mathrm{n}+4\frac{\left(\mathrm{n}-1\right)\mathrm{n}}{2}.\frac{1}{\mathrm{n}}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}\left(2\mathrm{n}-2+1\right)}{6}.\frac{1}{{\mathrm{n}}^{2}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \sum \mathrm{n}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2},\sum {\mathrm{n}}^{2}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[4+4\frac{\left(1-\frac{1}{\mathrm{n}}\right)}{2}+\frac{\left(1-\frac{1}{\mathrm{n}}\right)\left(2-\frac{1}{\mathrm{n}}\right)}{6}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left[4+4\frac{\left(1-0\right)}{2}+\frac{\left(1-0\right)\left(2-0\right)}{6}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4+2+\frac{1}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{19}{3}\end{array}$

Q.101

$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}.\phantom{\rule{0ex}{0ex}}{\int }_{1}^{4}\left({\mathrm{x}}^{2}-\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\int }_{1}^{4}\left({\mathrm{x}}^{2}-\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}={\int }_{1}^{4}{\mathrm{x}}^{2}\mathrm{dx}-{\int }_{1}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{I}}_{1}-{\mathrm{I}}_{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{Let}\right)\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{h}\right)+...+\mathrm{f}\left(\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{2},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{4-1}{\mathrm{n}}=\frac{3}{\mathrm{n}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{I}}_{1}={\int }_{1}^{4}{\mathrm{x}}^{2}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(4-1\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(1\right)+\mathrm{f}\left(1+\mathrm{h}\right)+...+\mathrm{f}\left(1+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[{\left(1\right)}^{2}+{\left(1+\mathrm{h}\right)}^{2}+...+{\left\{1+\left(\mathrm{n}-1\right)\mathrm{h}\right\}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[1+\left(1+2\mathrm{h}+{\mathrm{h}}^{2}\right)+...+\left\{1+2\left(\mathrm{n}-1\right)\mathrm{h}+{\left(\mathrm{n}-1\right)}^{2}{\mathrm{h}}^{2}\right\}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\left\{1+2+...+\left(\mathrm{n}-1\right)\right\}2\mathrm{h}+\left\{1+{2}^{2}+{3}^{2}+...+{\left(\mathrm{n}-1\right)}^{2}\right\}{\mathrm{h}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}}{2}2\mathrm{h}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}\left(2\mathrm{n}-2+1\right)}{6}{\mathrm{h}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\left(\mathrm{n}-1\right)\mathrm{n}.\frac{3}{\mathrm{n}}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}\left(2\mathrm{n}-2+1\right)}{6}.\frac{9}{{\mathrm{n}}^{2}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \sum \text{n}=\frac{\text{n}\left(\text{n}+1\right)}{2},\sum {\text{n}}^{2}=\frac{\text{n}\left(\text{n}+1\right)\left(2\text{n}+1\right)}{6}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[1+\left(1-\frac{1}{\mathrm{n}}\right)×3+\frac{\left(1-\frac{1}{\mathrm{n}}\right)\left(2-\frac{1}{\mathrm{n}}\right)}{6}×9\right]\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left[1+3\left(1-0\right)+\frac{3\left(1-0\right)\left(2-0\right)}{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=3\left(1+3+3\right)\\ \text{\hspace{0.17em}}=21\\ \mathrm{For}{\text{​\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}}_{\text{2}}={\int }_{1}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \therefore {\int }_{1}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\left(4-1\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(1\right)+\mathrm{f}\left(1+\mathrm{h}\right)+...+\mathrm{f}\left(1+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[1+1+\mathrm{h}+...+1+\left(\mathrm{n}-1\right)\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\mathrm{h}+2\mathrm{h}+...+\left(\mathrm{n}-1\right)\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\left\{1+2+...+\left(\mathrm{n}-1\right)\right\}\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{n}+\frac{\left(\mathrm{n}-1\right)\mathrm{n}}{2}\mathrm{h}\right]\left[\because \sum \mathrm{n}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[1+\frac{\left(\mathrm{n}-1\right)}{2}\left(\frac{3}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em} \hspace{0.17em}}=3\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[1+\frac{3}{2}\left(1-\frac{1}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left[1+\frac{3}{2}\left(1-0\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left(\frac{5}{2}\right)=\frac{15}{2}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}{\int }_{1}^{4}\left({\mathrm{x}}^{2}-\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{I}}_{1}-{\mathrm{I}}_{2}\text{\hspace{0.17em}}=21-\frac{15}{2}\text{\hspace{0.17em}}=\frac{27}{2}\end{array}$

Q.102

$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}.\phantom{\rule{0ex}{0ex}}{\int }_{–1}^{1}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em} have,}{\int }_{–1}^{1}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \because {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{h}\right)+...+\mathrm{f}\left(\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right],\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=-1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{1-\left(-1\right)}{\mathrm{n}}=\frac{2}{\mathrm{n}}\\ \therefore {\int }_{–1}^{1}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=\left(1+1\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(-1\right)+\mathrm{f}\left(-1+\mathrm{h}\right)+...+\mathrm{f}\left(-1+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[{\mathrm{e}}^{-1}+{\mathrm{e}}^{-1+\mathrm{h}}+...+{\mathrm{e}}^{-1+\left(\mathrm{n}-1\right)\mathrm{h}}\right]\\ \text{Using the sum to n terms of a G.P., where a}={\mathrm{e}}^{-1},\text{r}={\mathrm{e}}^{\mathrm{h}},\text{\hspace{0.17em}}\\ \mathrm{we}\text{have}\\ \therefore {\int }_{–1}^{1}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=2\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^{-1}\left({\mathrm{e}}^{\mathrm{nh}}-1\right)}{\left({\mathrm{e}}^{\mathrm{h}}-1\right)}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^{-1}\left({\mathrm{e}}^{\mathrm{n}.\frac{2}{\mathrm{n}}}-1\right)}{\left({\mathrm{e}}^{\frac{2}{\mathrm{n}}}-1\right)}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^{-1}\left({\mathrm{e}}^{2}-1\right)}{\left({\mathrm{e}}^{\frac{2}{\mathrm{n}}}-1\right)}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2{\mathrm{e}}^{-1}\left({\mathrm{e}}^{2}-1\right)}{\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[\frac{{\mathrm{e}}^{\frac{2}{\mathrm{n}}}-1}{\frac{2}{\mathrm{n}}}\right].2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2{\mathrm{e}}^{-1}\left({\mathrm{e}}^{2}-1\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{e}-\frac{1}{\mathrm{e}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \underset{\text{x}\to 0}{\mathrm{lim}}\frac{{\text{e}}^{\text{h}}-1}{\text{h}}=1\right]\end{array}$

Q.103

$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}.\phantom{\rule{0ex}{0ex}}{\int }_{0}^{4}\left(\mathrm{x}+{\mathrm{e}}^{2\mathrm{x}}\right)\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\int }_{0}^{4}\left(\mathrm{x}+{\mathrm{e}}^{2\mathrm{x}}\right)\text{\hspace{0.17em}}\mathrm{dx}={\int }_{0}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}+{\int }_{0}^{4}{\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{I}}_{1}+{\mathrm{I}}_{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{Let}\right)\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\text{\hspace{0.17em}}\mathrm{dx}=\left(\mathrm{b}-\mathrm{a}\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{h}\right)+...+\mathrm{f}\left(\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=0,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{4-0}{\mathrm{n}}=\frac{4}{\mathrm{n}}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}}_{\text{1}}={\int }_{0}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \therefore {\int }_{0}^{4}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\left(4-0\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(0\right)+\mathrm{f}\left(0+\mathrm{h}\right)+...+\mathrm{f}\left(0+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[0+\mathrm{h}+...+\left(\mathrm{n}-1\right)\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{h}+2\mathrm{h}+...+\left(\mathrm{n}-1\right)\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\left\{1+2+...+\left(\mathrm{n}-1\right)\right\}\mathrm{h}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\frac{\left(\mathrm{n}-1\right)\mathrm{n}}{2}\mathrm{h}\right]\left[\because \sum \mathrm{n}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[\frac{\left(\mathrm{n}-1\right)}{2}\left(\frac{4}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[\frac{4}{2}\left(1-\frac{1}{\mathrm{n}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\left[2\left(1-0\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\left(2\right)=8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{I}}_{2}={\int }_{0}^{4}{\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(4-0\right)\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\mathrm{f}\left(0\right)+\mathrm{f}\left(0+\mathrm{h}\right)+...+\mathrm{f}\left(0+\left(\mathrm{n}-1\right)\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[{\mathrm{e}}^{0}+{\mathrm{e}}^{2\mathrm{h}}+...+{\mathrm{e}}^{2\left(\mathrm{n}-1\right)\mathrm{h}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[1+{\mathrm{e}}^{2\mathrm{h}}+...+{\mathrm{e}}^{2\left(\mathrm{n}-1\right)\mathrm{h}}\right]\\ \text{Using the sum to n terms of a G.P., where a}=1,\text{r}={\mathrm{e}}^{2\mathrm{h}},\text{\hspace{0.17em}}\\ \mathrm{we}\text{have}\end{array}$

$\begin{array}{l}\therefore {\int }_{0}^{4}{\mathrm{e}}^{2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\frac{1.\left({\mathrm{e}}^{2\mathrm{nh}}-1\right)}{\left({\mathrm{e}}^{2\mathrm{h}}-1\right)}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\frac{\left({\mathrm{e}}^{2\mathrm{n}.\frac{4}{\mathrm{n}}}-1\right)}{\left({\mathrm{e}}^{\frac{8}{\mathrm{n}}}-1\right)}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\frac{1}{\mathrm{n}}\left[\frac{\left({\mathrm{e}}^{8}-1\right)}{\left({\mathrm{e}}^{\frac{8}{\mathrm{n}}}-1\right)}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4\left({\mathrm{e}}^{8}-1\right)}{\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{lim}}\left[\frac{{\mathrm{e}}^{\frac{8}{\mathrm{n}}}-1}{\frac{8}{\mathrm{n}}}\right].8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4\left({\mathrm{e}}^{8}-1\right)}{8}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{e}}^{8}-1}{2}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}{\int }_{0}^{4}\left(\mathrm{x}+{\mathrm{e}}^{2\mathrm{x}}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{I}}_{1}+{\mathrm{I}}_{2}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+\frac{{\mathrm{e}}^{8}-1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15+{\mathrm{e}}^{8}}{2}\end{array}$

Q.104

${\int }_{–1}^{1}\left(\mathrm{x}+1\right)\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{-1}^{1}\left(\mathrm{x}+1\right)\mathrm{dx}\\ \because \int \left(\mathrm{x}+1\right)\text{\hspace{0.17em}}\mathrm{dx}=\frac{{\mathrm{x}}^{2}}{2}+\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(-1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{{1}^{2}}{2}+1\right)-\left\{\frac{{\left(-1\right)}^{2}}{2}+\left(-1\right)\right\}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2}-\left(-\frac{1}{2}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\end{array}$

Q.105

${\int }_{2}^{3}\frac{1}{\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{2}^{3}\frac{1}{\mathrm{x}}\mathrm{dx}\\ \because \text{\hspace{0.17em}}\int \text{\hspace{0.17em}}\frac{1}{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}=\mathrm{logx}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(3\right)-\mathrm{F}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}3-\mathrm{log}2=\mathrm{log}\frac{3}{2}\end{array}$

Q.106

${\int }_{1}^{2}\left(4{\mathrm{x}}^{3}-5{\mathrm{x}}^{2}+6\mathrm{x}+9\right)\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{1}^{2}\left(4{\mathrm{x}}^{3}-5{\mathrm{x}}^{2}+6\mathrm{x}+9\right)\mathrm{dx}\\ \because \int \left(4{\mathrm{x}}^{3}-5{\mathrm{x}}^{2}+6\mathrm{x}+9\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int 4{\mathrm{x}}^{3}\text{\hspace{0.17em}}\mathrm{dx}-\int 5{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{dx}+\int 6\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}+\int 9\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{4}-\frac{5}{3}{\mathrm{x}}^{3}+3{\mathrm{x}}^{2}+9\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(2\right)-\mathrm{F}\left(1\right)\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{{2}^{4}-\frac{5}{3}{\left(2\right)}^{3}+3{\left(2\right)}^{2}+9\left(2\right)\right\}-\left\{{\left(1\right)}^{4}-\frac{5}{3}{\left(1\right)}^{3}+3{\left(1\right)}^{2}+9\left(1\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(16-\frac{40}{3}+12+18\right)-\left(1-\frac{5}{3}+3+9\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(46-\frac{40}{3}\right)-\left(13-\frac{5}{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=46-\frac{40}{3}-13+\frac{5}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=33-\frac{35}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{64}{3}\end{array}$

Q.107

${\int }_{0}^{\frac{\mathrm{\pi }}{4}}\mathrm{sin}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\frac{\mathrm{\pi }}{4}}\mathrm{sin}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \because \int \mathrm{sin}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=-\frac{1}{2}\mathrm{cos}2\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{2}\mathrm{cos}2\left(\frac{\mathrm{\pi }}{4}\right)-\left(-\frac{1}{2}\mathrm{cos}0\right)\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\mathrm{cos}\frac{\mathrm{\pi }}{2}+\frac{1}{2}\mathrm{cos}0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=0+\frac{1}{2}\left(1\right)=\frac{1}{2}\end{array}$

Q.108

${\int }_{0}^{\frac{\mathrm{\pi }}{2}}\mathrm{cos}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\mathrm{cos}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \because \int \mathrm{cos}2\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\mathrm{sin}2\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{2}\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\mathrm{sin}2\left(\frac{\mathrm{\pi }}{2}\right)-\left(-\frac{1}{2}\mathrm{sin}0\right)\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\mathrm{sin\pi }+\frac{1}{2}\mathrm{sin}0\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0+0=0\end{array}$

Q.109

${\int }_{4}^{5}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{4}^{5}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \because \int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}={\mathrm{e}}^{\mathrm{x}}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(5\right)-\mathrm{F}\left(4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{5}-{\mathrm{e}}^{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{4}\left(\mathrm{e}-1\right)\end{array}$

Q.110

${\int }_{0}^{\frac{\mathrm{\pi }}{4}}\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\frac{\mathrm{\pi }}{4}}\mathrm{tan}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \because \int \mathrm{tan}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=-\mathrm{log}|\mathrm{cosx}|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\mathrm{log}|\mathrm{cos}\frac{\mathrm{\pi }}{4}|-\left(-\mathrm{log}|\mathrm{cos}0|\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\mathrm{log}|\mathrm{cos}\frac{\mathrm{\pi }}{4}|+\mathrm{log}|\mathrm{cos}0|\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|\frac{1}{\sqrt{2}}|+\mathrm{log}1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{log}\sqrt{2}+0=\frac{1}{2}\mathrm{log}2\end{array}$

Q.111

${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{4}}\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{4}}\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}\\ \because \int \mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}=\mathrm{log}|\mathrm{cosecx}-\mathrm{cotx}|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(\frac{\mathrm{\pi }}{6}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\mathrm{cosec}\frac{\mathrm{\pi }}{4}-\mathrm{cot}\frac{\mathrm{\pi }}{4}|-\mathrm{log}|\mathrm{cosec}\frac{\mathrm{\pi }}{6}-\mathrm{cot}\frac{\mathrm{\pi }}{6}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\sqrt{2}-1|-\mathrm{log}|2-\sqrt{3}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\end{array}$

Q.112

${\int }_{0}^{1}\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{1}\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\mathrm{dx}\\ \because \int \frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\text{\hspace{0.17em}}\mathrm{dx}={\mathrm{sin}}^{-1}\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{sin}}^{-1}\left(1\right)-{\mathrm{sin}}^{-1}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\mathrm{\pi }}{2}-0=\frac{\mathrm{\pi }}{2}\end{array}$

Q.113

${\int }_{0}^{1}\frac{1}{\left(1+{\mathrm{x}}^{2}\right)}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{1}\frac{1}{1+{\mathrm{x}}^{2}}\mathrm{dx}\\ \because \int \frac{1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}={\mathrm{tan}}^{-1}\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\mathrm{tan}}^{-1}\left(1\right)-{\mathrm{tan}}^{-1}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\mathrm{\pi }}{4}-0=\frac{\mathrm{\pi }}{4}\end{array}$

Q.114

${\int }_{2}^{3}\frac{\mathrm{dx}}{\left({\mathrm{x}}^{2}-1\right)}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{2}^{3}\frac{\mathrm{dx}}{\left({\mathrm{x}}^{2}-1\right)}\\ \because \int \frac{1}{{\mathrm{x}}^{2}-1}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\mathrm{log}|\frac{\mathrm{x}-1}{\mathrm{x}+1}|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(3\right)-\mathrm{F}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\frac{3-1}{3+1}|-\frac{1}{2}\mathrm{log}|\frac{2-1}{2+1}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\frac{2}{4}|-\frac{1}{2}\mathrm{log}|\frac{1}{3}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\frac{2}{4}×\frac{3}{1}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\frac{3}{2}|\end{array}$

Q.115

${\int }_{0}^{\frac{\mathrm{\pi }}{2}}{\mathrm{cos}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}{\mathrm{cos}}^{2}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{1+\mathrm{cos}2\mathrm{x}}{2}\text{\hspace{0.17em}}\mathrm{dx}\\ \because \int \frac{1+\mathrm{cos}2\mathrm{x}}{2}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\left(\mathrm{x}+\frac{\mathrm{sin}2\mathrm{x}}{2}\right)=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{2}\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left\{\frac{\mathrm{\pi }}{2}+\frac{\mathrm{sin}2.\left(\frac{\mathrm{\pi }}{2}\right)}{2}\right\}-\frac{1}{2}\left\{0+\frac{\mathrm{sin}2\left(0\right)}{2}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{sin\pi }}{2}\right)-\frac{1}{2}\left\{0+0\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{4}\end{array}$

Q.116

${\int }_{2}^{3}\frac{\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}}{\left({\mathrm{x}}^{2}+1\right)}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{2}^{3}\frac{\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}}{\left({\mathrm{x}}^{2}+1\right)}\\ \because \int \frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\mathrm{log}|{\mathrm{x}}^{2}+1|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(3\right)-\mathrm{F}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|{3}^{2}+1|-\frac{1}{2}\mathrm{log}|{2}^{2}+1|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|10|-\frac{1}{2}\mathrm{log}|5|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\frac{10}{5}|\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|2|\end{array}$

Q.117

${\int }_{0}^{1}\frac{2\mathrm{x}+3}{\left(5{\mathrm{x}}^{2}+1\right)}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{1}\frac{2\mathrm{x}+3}{\left(5{\mathrm{x}}^{2}+1\right)}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}{\int }_{0}^{1}\frac{10\mathrm{x}}{\left(5{\mathrm{x}}^{2}+1\right)}\mathrm{dx}+{\int }_{0}^{1}\frac{3}{\left(5{\mathrm{x}}^{2}+1\right)}\mathrm{dx}\\ \int \frac{2\mathrm{x}+3}{\left(5{\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{5}\int \frac{10\mathrm{x}}{\left(5{\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}+3\int \frac{1}{\left(5{\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\mathrm{log}|5{\mathrm{x}}^{2}+1|+3\int \frac{1}{{\left(\sqrt{5}\mathrm{x}\right)}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\mathrm{log}|5{\mathrm{x}}^{2}+1|+3{\mathrm{tan}}^{-1}\sqrt{5}\mathrm{x}.\frac{1}{\sqrt{5}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\mathrm{log}|5{\mathrm{x}}^{2}+1|+\frac{3}{\sqrt{5}}{\mathrm{tan}}^{-1}\left(\sqrt{5}\mathrm{x}\right)=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\mathrm{log}|5{\left(1\right)}^{2}+1|+\frac{3}{\sqrt{5}}{\mathrm{tan}}^{-1}\left\{\sqrt{5}\left(1\right)\right\}-\left[\begin{array}{l}\frac{1}{5}\mathrm{log}|5{\left(0\right)}^{2}+1|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\frac{3}{\sqrt{5}}{\mathrm{tan}}^{-1}\left\{\sqrt{5}\left(0\right)\right\}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\mathrm{log}\left(6\right)+\frac{3}{\sqrt{5}}{\mathrm{tan}}^{-1}\left(\sqrt{5}\right)-\frac{1}{5}\mathrm{log}|1|-\frac{3}{\sqrt{5}}{\mathrm{tan}}^{-1}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\mathrm{log}\left(6\right)+\frac{3}{\sqrt{5}}{\mathrm{tan}}^{-1}\left(\sqrt{5}\right)\end{array}$

Q.118

${\int }_{0}^{1}\mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{{\mathrm{x}}^{2}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{1}\mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{{\mathrm{x}}^{2}}\mathrm{dx}\\ \because \int \mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{t}}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{2\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Let​ t}={\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}{\mathrm{e}}^{\mathrm{t}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^{2}}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}{\mathrm{e}}^{{1}^{2}}-\frac{1}{2}{\mathrm{e}}^{0}\text{\hspace{0.17em}}=\frac{1}{2}\left(\mathrm{e}-1\right)\end{array}$

Q.119

${\int }_{1}^{2}\frac{5{\mathrm{x}}^{2}}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}={\int }_{1}^{2}\frac{5{\mathrm{x}}^{2}}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}\\ \int \frac{5{\mathrm{x}}^{2}}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}=\int \left(5-\frac{20\mathrm{x}+15}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int 5\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{20\mathrm{x}+15}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20\mathrm{x}+15=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}+4\mathrm{x}+3\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{A}\left(2\mathrm{x}+4\right)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{Ax}+4\mathrm{A}+\mathrm{B}\\ \text{Equating the coefficients of x and constant term, we get}\\ ⇒\text{\hspace{0.17em}}2\mathrm{A}=20\text{\hspace{0.17em}and\hspace{0.17em}4A}+\text{B}=\text{15}\\ ⇒\text{\hspace{0.17em}}\mathrm{A}=10\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=-25\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20\mathrm{x}+15=10\left(2\mathrm{x}+4\right)-25\\ \text{From equation}\left(\text{i}\right)\text{, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}}\int \frac{5{\mathrm{x}}^{2}}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}=\int 5\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{10\left(2\mathrm{x}+4\right)-25}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int 5\text{\hspace{0.17em}}\mathrm{dx}-10\int \frac{\left(2\mathrm{x}+4\right)}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}+25\int \frac{1}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\mathrm{x}-10\mathrm{log}\left({\mathrm{x}}^{2}+4\mathrm{x}+3\right)+25\int \frac{1}{{\left(\mathrm{x}+2\right)}^{2}-1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\mathrm{x}-10\mathrm{log}\left({\mathrm{x}}^{2}+4\mathrm{x}+3\right)+25×\frac{1}{2}\mathrm{log}|\frac{\mathrm{x}+2-1}{\mathrm{x}+2+1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\mathrm{x}-10\mathrm{log}\left({\mathrm{x}}^{2}+4\mathrm{x}+3\right)+\frac{25}{2}\mathrm{log}|\frac{\mathrm{x}+1}{\mathrm{x}+3}|\\ {\int }_{1}^{2}\frac{5{\mathrm{x}}^{2}}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}={{\left[5\mathrm{x}-10\mathrm{log}\left({\mathrm{x}}^{2}+4\mathrm{x}+3\right)+\frac{25}{2}\mathrm{log}|\frac{\mathrm{x}+1}{\mathrm{x}+3}|\right]}_{1}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left[5\left(2\right)-10\mathrm{log}\left({2}^{2}+4×2+3\right)+\frac{25}{2}\mathrm{log}|\frac{2+1}{2+3}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}-\left[5\left(1\right)-10\mathrm{log}\left({1}^{2}+4×1+3\right)+\frac{25}{2}\mathrm{log}|\frac{1+1}{1+3}|\right]\\ {\int }_{1}^{2}\frac{5{\mathrm{x}}^{2}}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}={{\left[5\mathrm{x}-10\mathrm{log}\left({\mathrm{x}}^{2}+4\mathrm{x}+3\right)+\frac{25}{2}\mathrm{log}|\frac{\mathrm{x}+1}{\mathrm{x}+3}|\right]}_{1}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left[5\left(2\right)-10\mathrm{log}\left({2}^{2}+4×2+3\right)+\frac{25}{2}\mathrm{log}|\frac{2+1}{2+3}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\left[5\left(1\right)-10\mathrm{log}\left({1}^{2}+4×1+3\right)+\frac{25}{2}\mathrm{log}|\frac{1+1}{1+3}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(10-10\mathrm{log}15+\frac{25}{2}\mathrm{log}\frac{3}{5}\right)-\left(5-10\mathrm{log}8+\frac{25}{2}\mathrm{log}\frac{2}{4}\right)\\ \text{\hspace{0.17em}}=5-10\left(\mathrm{log}15-\mathrm{log}8\right)+\frac{25}{2}\left(\mathrm{log}\frac{3}{5}-\mathrm{log}\frac{2}{4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5-10\left(\mathrm{log}5+\mathrm{log}3-\mathrm{log}4-\mathrm{log}2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\frac{25}{2}\left(\mathrm{log}3-\mathrm{log}5-\mathrm{log}2+\mathrm{log}4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5-\left(10+\frac{25}{2}\right)\mathrm{log}5+\left(10+\frac{25}{2}\right)\mathrm{log}4+\left(-10+\frac{25}{2}\right)\mathrm{log}3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\left(10-\frac{25}{2}\right)\mathrm{log}2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5-\frac{45}{2}\mathrm{log}5+\frac{45}{2}\mathrm{log}4+\frac{5}{2}\mathrm{log}3-\frac{5}{2}\mathrm{log}2\\ {\int }_{1}^{2}\frac{5{\mathrm{x}}^{2}}{{\mathrm{x}}^{2}+4\mathrm{x}+3}\text{\hspace{0.17em}}\mathrm{dx}\text{\hspace{0.17em}}=5-\frac{45}{2}\mathrm{log}\frac{5}{4}+\frac{5}{2}\mathrm{log}\frac{3}{2}\text{\hspace{0.17em}\hspace{0.17em}}=5-\frac{5}{2}\left(9\mathrm{log}\frac{5}{4}-\mathrm{log}\frac{3}{2}\right)\end{array}$

Q.120

${\int }_{0}^{\frac{\mathrm{\pi }}{4}}\left(2{\mathrm{sec}}^{2}\text{\hspace{0.17em}}\mathrm{x}+{\mathrm{x}}^{3}+2\right)\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\frac{\mathrm{\pi }}{4}}\left(2{\mathrm{sec}}^{2}\text{\hspace{0.17em}}\mathrm{x}+{\mathrm{x}}^{3}+2\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \int \left(2{\mathrm{sec}}^{2}\text{\hspace{0.17em}}\mathrm{x}+{\mathrm{x}}^{3}+2\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\int {\mathrm{sec}}^{2}\text{\hspace{0.17em}}\mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}+\int {\mathrm{x}}^{3}\text{\hspace{0.17em}}\mathrm{dx}+\int 2\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{tanx}+\frac{{\mathrm{x}}^{4}}{4}+2\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\left\{2\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}\right)+\frac{{\left(\frac{\mathrm{\pi }}{4}\right)}^{4}}{4}+2\left(\frac{\mathrm{\pi }}{4}\right)\right\}-\left\{2\mathrm{tan}\left(0\right)+\frac{{\left(0\right)}^{4}}{4}+2\left(0\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=2\left[\mathrm{tan}\frac{\mathrm{\pi }}{4}-\mathrm{tan}0\right]+\frac{1}{4}\left[{\left(\frac{\mathrm{\pi }}{4}\right)}^{4}-0\right]+2\left[\frac{\mathrm{\pi }}{4}-0\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=2\left[1-0\right]+\frac{1}{4}\left[{\left(\frac{\mathrm{\pi }}{4}\right)}^{4}\right]+2\left[\frac{\mathrm{\pi }}{4}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=2+\frac{{\mathrm{\pi }}^{4}}{1024}+\frac{\mathrm{\pi }}{2}\end{array}$

Q.121

${\int }_{0}^{\mathrm{\pi }}\left({\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}-{\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}\right)\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\mathrm{\pi }}\left({\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}-{\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}\right)\mathrm{dx}\\ \int \left({\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}-{\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}=-\int \left({\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}-{\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}\right)\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\int \mathrm{cosx}\text{\hspace{0.17em}}\mathrm{dx}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because {\mathrm{cos}}^{2}\mathrm{\theta }-{\mathrm{sin}}^{2}\mathrm{\theta }=\mathrm{cos}2\mathrm{\theta }\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{sinx}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\mathrm{\pi }\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{sin\pi }-\left(-\mathrm{sin}0\right)\text{\hspace{0.17em}\hspace{0.17em}}=0\end{array}$

Q.122

${\int }_{0}^{2}\frac{6\mathrm{x}+3}{{\mathrm{x}}^{2}+4}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{2}\frac{6\mathrm{x}+3}{{\mathrm{x}}^{2}+4}\mathrm{dx}\\ \int \frac{6\mathrm{x}+3}{{\mathrm{x}}^{2}+4}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{6\mathrm{x}}{{\mathrm{x}}^{2}+4}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{3}{{\mathrm{x}}^{2}+4}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=3\int \frac{2\mathrm{x}}{{\mathrm{x}}^{2}+4}\text{\hspace{0.17em}}\mathrm{dx}+3\int \frac{1}{{\mathrm{x}}^{2}+{2}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left[\mathrm{log}\left({\mathrm{x}}^{2}+4\right)\right]+\frac{3}{2}\left[{\mathrm{tan}}^{-1}\frac{\mathrm{x}}{2}\right]=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(2\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left[\mathrm{log}\left({2}^{2}+4\right)-\frac{3}{2}{\mathrm{tan}}^{-1}\frac{2}{2}\right]-\left[\mathrm{log}\left({0}^{2}+4\right)+\frac{3}{2}{\mathrm{tan}}^{-1}\frac{0}{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left[\mathrm{log}8-\mathrm{log}4\right]-\frac{3}{2}\left[{\mathrm{tan}}^{-1}\left(1\right)-{\mathrm{tan}}^{-1}\left(0\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{log}\frac{8}{4}-\frac{3}{2}\left[\frac{\mathrm{\pi }}{4}-0\right]\\ {\int }_{0}^{2}\frac{6\mathrm{x}+3}{{\mathrm{x}}^{2}+4}\mathrm{dx}=3\mathrm{log}2-\frac{3\mathrm{\pi }}{8}\end{array}$

Q.123

${\int }_{0}^{1}\left(\mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}+\mathrm{sin}\frac{\mathrm{\pi x}}{4}\right)\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{1}\left(\mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}+\mathrm{sin}\frac{\mathrm{\pi x}}{4}\right)\mathrm{dx}\\ \int \left(\mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}+\mathrm{sin}\frac{\mathrm{\pi x}}{4}\right)\text{\hspace{0.17em}}\mathrm{dx}=\int \mathrm{x}\text{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}+\int \mathrm{sin}\frac{\mathrm{\pi x}}{4}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{x}\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\int \left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\right)\text{\hspace{0.17em}}\mathrm{dx}-\frac{1}{\left(\frac{\mathrm{\pi }}{4}\right)}\mathrm{cos}\frac{\mathrm{\pi x}}{4}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{xe}}^{\mathrm{x}}-\int {\mathrm{e}}^{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\frac{4}{\mathrm{\pi }}\mathrm{cos}\frac{\mathrm{\pi x}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{xe}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}-\frac{4}{\mathrm{\pi }}\mathrm{cos}\frac{\mathrm{\pi x}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}\left(\mathrm{x}-1\right)-\frac{4}{\mathrm{\pi }}\mathrm{cos}\frac{\mathrm{\pi x}}{4}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{{\mathrm{e}}^{1}\left(1-1\right)-\frac{4}{\mathrm{\pi }}\mathrm{cos}\frac{\mathrm{\pi }\left(1\right)}{4}\right\}-\left\{{\mathrm{e}}^{0}\left(0-1\right)-\frac{4}{\mathrm{\pi }}\mathrm{cos}\frac{\mathrm{\pi }\left(0\right)}{4}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{-\frac{4}{\mathrm{\pi }}\mathrm{cos}\frac{\mathrm{\pi }}{4}\right\}-\left\{-1-\frac{4}{\mathrm{\pi }}×1\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{-\frac{4}{\mathrm{\pi }}×\frac{1}{\sqrt{2}}\right\}-\left\{-1-\frac{4}{\mathrm{\pi }}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{4}{\sqrt{2}\mathrm{\pi }}+1+\frac{4}{\mathrm{\pi }}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1+\frac{4}{\mathrm{\pi }}-\frac{2\sqrt{2}}{\mathrm{\pi }}\end{array}$

Q.124

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ {\int }_{1}^{\sqrt{3}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{2}}\text{equals}\\ \left(\text{A}\right)\text{ }\frac{\mathrm{\pi }}{3}\\ \left(\text{B}\right)\text{ }\frac{2\mathrm{\pi }}{3}\\ \left(\text{C}\right)\text{ }\frac{\mathrm{\pi }}{6}\\ \left(\text{D}\right)\text{ }\frac{\mathrm{\pi }}{12}\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{1}^{\sqrt{3}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{2}}\\ \int \frac{1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}={\mathrm{tan}}^{-1}\mathrm{x}\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\sqrt{3}\right)-\mathrm{F}\left(1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{tan}}^{-1}\left(\sqrt{3}\right)-{\text{tan}}^{-1}\left(1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4\mathrm{\pi }-3\mathrm{\pi }}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{\pi }}{12}\\ \text{Hence, the correct option is D.}\end{array}$

Q.125

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ {\int }_{0}^{\frac{2}{3}}\frac{\mathrm{dx}}{4+9{\mathrm{x}}^{2}}\text{ equals}\\ \left(\text{A}\right)\text{ }\frac{\mathrm{\pi }}{6}\\ \left(\text{B}\right)\text{ }\frac{\mathrm{\pi }}{12}\\ \left(\text{C}\right)\text{ }\frac{\mathrm{\pi }}{24}\\ \left(\text{D}\right)\text{ }\frac{\mathrm{\pi }}{4}\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\frac{2}{3}}\frac{\mathrm{dx}}{4+9{\mathrm{x}}^{2}}\\ \int \frac{1}{4+9{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{{2}^{2}+{\left(3\mathrm{x}\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}{\mathrm{tan}}^{-1}\left(\frac{3\mathrm{x}}{2}\right)×\frac{1}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}{\mathrm{tan}}^{-1}\left(\frac{3\mathrm{x}}{2}\right)=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{2}{3}\right)-\mathrm{F}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}{\mathrm{tan}}^{-1}\left(\frac{3}{2}×\frac{2}{3}\right)-\frac{1}{6}{\mathrm{tan}}^{-1}\left(\frac{3}{2}×0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}{\mathrm{tan}}^{-1}\left(1\right)-\frac{1}{6}{\mathrm{tan}}^{-1}\left(0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}×\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{24}\\ \text{Hence, the correct option is C.}\end{array}$

Q.126

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_{0}^{1}\frac{\mathrm{x}}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}$

Ans.

$\begin{array}{l}{\int }_{0}^{1}\frac{\mathrm{x}}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{2}+1⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\\ \text{When x}=0,\text{\hspace{0.17em}t}=1\text{and when x}=\text{1, t}=\text{2}\\ \therefore {\int }_{0}^{1}\frac{\mathrm{x}}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}{\int }_{1}^{2}\frac{1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}{{\left[\mathrm{log}|\mathrm{t}|\right]}_{1}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[\mathrm{log}2-\mathrm{log}1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}2\end{array}$

Q.127

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_{0}^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\mathrm{cos}}^{5}\mathrm{\varphi }\text{\hspace{0.17em}}\mathrm{d\varphi }$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\mathrm{cos}}^{5}\mathrm{\varphi }\text{\hspace{0.17em}}\mathrm{d}\mathrm{\varphi }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\mathrm{cos}}^{4}\mathrm{\varphi }\text{\hspace{0.17em}}\mathrm{cos\varphi }\text{\hspace{0.17em}}\mathrm{d\varphi }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\left({\mathrm{cos}}^{2}\mathrm{\varphi }\right)}^{2}\text{\hspace{0.17em}}\mathrm{cos\varphi }\text{\hspace{0.17em}}\mathrm{d\varphi }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\left(1-{\mathrm{sin}}^{2}\mathrm{\varphi }\right)}^{2}\text{\hspace{0.17em}}\mathrm{cos\varphi }\text{\hspace{0.17em}}\mathrm{d\varphi }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}\left(1-2{\mathrm{sin}}^{2}\mathrm{\varphi }+{\mathrm{sin}}^{4}\mathrm{\varphi }\right)\text{\hspace{0.17em}}\mathrm{cos\varphi }\text{\hspace{0.17em}}\mathrm{d\varphi }\\ \text{Let t}=\mathrm{sin\varphi }⇒\frac{\mathrm{dt}}{\mathrm{d\varphi }}=\mathrm{cos\varphi }\\ \text{When\hspace{0.17em}}\mathrm{\varphi }=0,\text{\hspace{0.17em}}\mathrm{t}=0\text{and when}\mathrm{\varphi }=\frac{\mathrm{\pi }}{2},\text{\hspace{0.17em}}\mathrm{t}=1\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{0}^{1}\sqrt{\mathrm{t}}\left(1-2{\mathrm{t}}^{2}+{\mathrm{t}}^{4}\right)\text{\hspace{0.17em}}\mathrm{dt}\\ ={\int }_{0}^{1}\left({\mathrm{t}}^{\frac{1}{2}}-2{\mathrm{t}}^{\frac{}{}}\end{array}$