What is the total number of questions in the NCERT Solutions for Class 12 Mathematics Chapter 7 Integrals?

NCERT Solutions Class 12 Mathematics Chapter 7 Integrals has 11 exercises with 227 problems. There are 150 basic and easy-to-solve problems among the 227 total, while the other 77 are moderate to complex long-answer questions that require more fundamental idea practice. In addition, there is one random exercise with 44 questions in this chapter to help students better understand the concept of Integrals.

What is the significance of NCERT Solutions for Class 12 Mathematics Chapter 7?

Extramarks has carefully designed NCERT Solutions for Class 12 Mathematics Chapter 7 after extensive research to provide a clear understanding of all fundamentals. Students will be able to achieve full conceptual knowledge and perform well in tests if they strategically prepare all of the relevant topics with the help of these resources. The well-crafted format of these solutions is ideal for encouraging arithmetic practice, which in turn boosts students’ confidence.

Is it necessary for me to practice all NCERT Solutions Class 12 Mathematics Chapter 7 Integrals?

Integral is a broad topic that necessitates a great deal of practice and the incorporation of newer concepts into previously studied topics. Students will have a steady grasp of subsequent topics by practising all the problems in NCERT Solutions Class 12 Mathematics Chapter 7 Integers. The sums in these answers have been carefully placed to provide step-by-step learning for responding to the chapter’s rising complexity. It will also help students develop the right technique for successfully handling complex challenges by improving their problem-solving abilities.

Why should I practise NCERT Solutions for Class 12 Mathematics Chapter 7 Integrals?

Students can quickly and steadily create a strong mathematical foundation for calculating sums related to integral calculus by practising the NCERT Solutions Class 12 Mathematics Integrals Chapter 7 regularly. These solutions will show them how to solve difficult questions asked in board and competitive exams. Furthermore, by reviewing these solutions, students can develop a strong conceptual fluency that will benefit them in the future.

NCERT Solutions Class 12 Mathematics Chapter 7

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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.

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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 Solutions – 22 Questions – Class 12 Mathematics

Chapter 7 Ex 7.2 Solutions – 39 Questions – Class 12 Mathematics

Chapter 7 Ex 7.3 Solutions – 24 Questions – Class 12 Mathematics

Chapter 7 Ex 7.4 Solutions – 25 Questions – Class 12 Mathematics

Chapter 7 Ex 7.5 Solutions – 23 Questions – Class 12 Mathematics

Chapter 7 Ex 7.6 Solutions – 24 Questions – Class 12 Mathematics

Chapter 7 Ex 7.7 Solutions – 11 Questions – Class 12 Mathematics

Chapter 7 Ex 7.8 Solutions – 6 Questions – Class 12 Mathematics

Chapter 7 Ex 7.9 Solutions – 22 Questions – Class 12 Mathematics

Chapter 7 Ex 7.10 Solutions – 10 Questions – Class 12 Mathematics

Chapter 7 Ex 7.11 Solutions – 21 Questions – Class 12 Mathematics

Chapter 7 Miscellaneous Exercise – 44 Questions – Class 12 Mathematics

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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

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

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

Ans.

$\begin{array}{l} We have, \\ \frac{cosx - sinx}{1 + \sin 2 x} = \frac{cosx - sinx}{\sin^{2} x + \cos^{2} x + 2 sinxcosx} \\ = \frac{cosx - sinx}{{(sinx + cosx)}^{2}} \\ ∴ \int \frac{cosx - sinx}{1 + \sin 2 x} dx \\ = \int \frac{cosx - sinx}{{(sinx + cosx)}^{2}} dx \\ Let t = sinx + cosx \Rightarrow \frac{dt}{dx} = cosx - sinx \\ \Rightarrow dx = \frac{dt}{cosx - sinx} \\ \int \frac{cosx - sinx}{1 + \sin 2 x} dx = \int \frac{cosx - sinx}{{(t)}^{2}} \frac{dt}{cosx - sinx} \\ = \int t^{- 2} dt \\ = - \frac{1}{t} + C \\ = - \frac{1}{sinx + cosx} + C \end{array}$

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

Ans.

$\begin{array}{l} \int \tan^{3} 2 xsec 2 x dx = \int \tan^{2} 2 xtan 2 xsec 2 x dx \\ = \int \tan^{2} 2 xtan 2 xsec 2 x dx \\ = \int (\sec^{2} 2 x - 1) \tan 2 xsec 2 x dx \\ = \int \sec^{2} 2 x . \tan 2 xsec 2 x dx - \int \tan 2 xsec 2 x dx \\ Let t = \sec 2 x \Rightarrow \frac{dt}{dx} = 2 \sec 2 xtan 2 x \\ \Rightarrow dx = \frac{dt}{2 \sec 2 xtan 2 x} \\ So, \\ \int \tan^{3} 2 xsec 2 x dx = \int t^{2} . \tan 2 xsec 2 x . \frac{dt}{2 \sec 2 xtan 2 x} - \frac{1}{2} \sec 2 x + C \\ = \frac{1}{2} \int t^{2} dt - \frac{1}{2} \sec 2 x + C \\ = \frac{1}{2} \times \frac{t^{3}}{3} - \frac{1}{2} \sec 2 x + C \\ = \frac{1}{6} \sec^{3} 2 x - \frac{1}{2} \sec 2 x + C \end{array}$

Q.3 Find the integral of the function tan⁴x.

Ans.

$\begin{array}{l} \int \tan^{4} x dx = \int \tan^{2} {xtan}^{2} x dx \\ = \int (\sec^{2} x - 1) \tan^{2} x dx \\ = \int \sec^{2} x \tan^{2} x dx - \int \tan^{2} x dx \\ = \int \sec^{2} x \tan^{2} x dx - \int (\sec^{2} x - 1) dx \\ = \int \sec^{2} x \tan^{2} x dx - \int \sec^{2} x dx + \int 1 dx \\ Let t = tanx \Rightarrow \frac{dt}{dx} = \sec^{2} x \Rightarrow dx = \frac{dt}{\sec^{2} x} \\ \int \tan^{4} x dx = \int \sec^{2} x . t^{2} \frac{dt}{\sec^{2} x} - tanx + x + C \\ = \int t^{2} dt - tanx + x + C \\ = \frac{t^{3}}{3} - tanx + x + C \\ = \frac{\tan^{3} x}{3} - tanx + x + C \end{array}$

Q.4

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

Ans.

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

Q.5

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

Ans.

$\begin{array}{l} We have, \\ \frac{\cos 2 x + 2 \sin^{2} x}{\cos^{2} x} = \frac{\cos 2 x + (1 - \cos 2 x)}{\cos^{2} x} \\ = \frac{1}{\cos^{2} x} \\ = \sec^{2} x \\ ∴ \int \frac{\cos 2 x + 2 \sin^{2} x}{\cos^{2} x} dx = \int \sec^{2} x dx \\ = tanx + C \end{array}$

Q.6

$Find the integral of the function \frac{1}{{sinxcos}^{3} x} .$

Ans.

$\begin{array}{l} We have, \frac{1}{sinx \cos^{3} x} = \frac{\sin^{2} x + \cos^{2} x}{{sinxcos}^{3} x} \\ = \frac{\sin^{2} x}{{sinxcos}^{3} x} + \frac{\cos^{2} x}{{sinxcos}^{3} x} \\ = tanx \sec^{2} x + \frac{1}{sinx cosx} \\ = tanx \sec^{2} x + \frac{\sec^{2} x}{tanx} \\ ∴ \int \frac{1}{{sinxcos}^{3} x} dx = \int tanx \sec^{2} x dx + \int \frac{\sec^{2} x}{tanx} dx \\ Let t = tanx \Rightarrow \frac{dt}{dx} = \sec^{2} x \Rightarrow dx = \frac{dt}{\sec^{2} x} \\ So, \int \frac{1}{sinx \cos^{3} x} dx = \int t \sec^{2} x . \frac{dt}{\sec^{2} x} + \int \frac{\sec^{2} x}{t} . \frac{dt}{\sec^{2} x} \\ = \int t dt + \int \frac{1}{t} dt \\ = \frac{t^{2}}{2} + \log |t| + C \\ ∴ \int \frac{1}{{sinxcos}^{3} x} dx = \frac{\tan^{2} x}{2} + \log |tanx| + C \end{array}$

Q.7

$Find the integral of the function given below . \frac{\cos 2 x}{{(cosx + sinx)}^{2}}$

Ans.

$\begin{array}{l} We have, \\ \frac{\cos 2 x}{{(cosx + sinx)}^{2}} = \frac{\cos 2 x}{(\cos^{2} x + \sin^{2} x + 2 sinxcosx)} \\ = \frac{\cos 2 x}{(1 + \sin 2 x)} \\ ∴ \int \frac{\cos 2 x}{{(cosx + sinx)}^{2}} dx = \int \frac{\cos 2 x}{(1 + \sin 2 x)} dx \\ Let t = 1 + \sin 2 x \Rightarrow \frac{dt}{dx} = 2 \cos 2 x \Rightarrow dx = \frac{dt}{2 \cos 2 x} \\ \Rightarrow \int \frac{\cos 2 x}{{(cosx + sinx)}^{2}} dx = \int \frac{\cos 2 x}{t} . \frac{dt}{2 \cos 2 x} \\ = \frac{1}{2} \int \frac{1}{t} dt \\ = \frac{1}{2} \log |t| + C \\ = \frac{1}{2} \log |1 + \sin 2 x| + C \\ = \frac{1}{2} \log |{(cosx + sinx)}^{2}| + C \\ = \frac{1}{2} \times 2 \log (cosx + sinx) + C \\ Hence, \int \frac{\cos 2 x}{{(cosx + sinx)}^{2}} dx = \log (cosx + sinx) + C \end{array}$

Q.8

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

Ans.

$\begin{array}{l} We have, \\ \int \sin^{- 1} (cosx) dx \\ Let t = cosx \Rightarrow sinx = \sqrt{1 - t^{2}} \\ and \frac{dt}{dx} = - sinx \\ \Rightarrow dx = \frac{dt}{- sinx} = \frac{- dt}{\sqrt{1 - t^{2}}} \\ ∴ \int \sin^{- 1} (cosx) dx = \int \sin^{- 1} t \times \frac{- dt}{\sqrt{1 - t^{2}}} \\ = - \int \frac{\sin^{- 1} t}{\sqrt{1 - t^{2}}} dt \\ = - \int \frac{u}{\sqrt{1 - t^{2}}} \sqrt{1 - t^{2}} du [\begin{array}{l} Let u = \sin^{- 1} t \\ \Rightarrow \frac{du}{dx} = \frac{1}{\sqrt{1 - t^{2}}} \\ dx = \sqrt{1 - t^{2}} du \end{array}] \\ = - \int u du \\ = - \frac{u^{2}}{2} + C \\ = - \frac{{(\sin^{- 1} t)}^{2}}{2} + C \\ \int \sin^{- 1} (cosx) dx = - \frac{{\{\sin^{- 1} (cosx)\}}^{2}}{2} + C \\ = - \frac{{\{\frac{π}{2} - \cos^{- 1} (cosx)\}}^{2}}{2} + C [\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}] \\ = - \frac{{\{\frac{π}{2} - x\}}^{2}}{2} + C \\ = - \frac{1}{2} (\frac{π^{2}}{4} - πx + x^{2}) + C \\ = - \frac{π^{2}}{8} + \frac{πx}{2} - \frac{x^{2}}{2} + C \\ = - \frac{x^{2}}{2} + \frac{πx}{2} + (C - \frac{π^{2}}{8}) \\ = \frac{πx}{2} - \frac{x^{2}}{2} + C ‘, where C’= C + \frac{π^{2}}{8} \end{array}$

Q.9

$Find the integral of the function given below . \frac{1}{\cos (x - a) \cos (x - b)}$

Ans.

$\begin{array}{l} We have \frac{1}{\cos (x - a) \cos (x - b)} \\ = \frac{1}{\sin (a - b)} [\frac{\sin (a - b)}{\cos (x - a) \cos (x - b)}] \\ = \frac{1}{\sin (a - b)} [\frac{\sin (x - b - x + a)}{\cos (x - a) \cos (x - b)}] \\ = \frac{1}{\sin (a - b)} [\frac{\sin \{(x - b) - (x - a)\}}{\cos (x - a) \cos (x - b)}] \\ = \frac{1}{\sin (a - b)} \{\frac{\sin (x - b) \cos (x - a) - \cos (x - b) \sin (x - a)}{\cos (x - a) \cos (x - b)}\} \\ = \frac{1}{\sin (a - b)} \{\frac{\sin (x - b) \cos (x - a)}{\cos (x - a) \cos (x - b)} - \frac{\cos (x - b) \sin (x - a)}{\cos (x - a) \cos (x - b)}\} \\ = \frac{1}{\sin (a - b)} \{\frac{\sin (x - b)}{\cos (x - b)} - \frac{\sin (x - a)}{\cos (x - a)}\} \\ = \frac{1}{\sin (a - b)} \{\tan (x - b) - \tan (x - a)\} \\ ∴ \int \frac{1}{\cos (x - a) \cos (x - b)} dx \\ = \frac{1}{\sin (a - b)} \{\int \tan (x - b) dx - \int \tan (x - a) dx\} \\ = \frac{1}{\sin (a - b)} \{- \log |\cos (x - b)| + \log |\cos (x - a)|\} + C \\ = \frac{1}{\sin (a - b)} \log |\frac{\cos (x - a)}{\cos (x - b)}| + C \end{array}$

Q.10

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

Ans.

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

Q.11

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

Ans.

$\begin{array}{l} We have \int \frac{e^{x} (1 + x)}{\cos^{2} (e^{x} x)} dx \\ Let t = e^{x} x \\ \Rightarrow \frac{dt}{dx} = e^{x} \frac{d}{dx} x + x \frac{d}{dx} e^{x} \\ = e^{x} + {xe}^{x} \\ = e^{x} (1 + x) \\ \Rightarrow dx = \frac{dt}{e^{x} (1 + x)} \\ ∴ \int \frac{e^{x} (1 + x)}{\cos^{2} (e^{x} x)} dx = \int \frac{e^{x} (1 + x)}{\cos^{2} t} \frac{dt}{e^{x} (1 + x)} \\ = \int \frac{1}{\cos^{2} t} dt \\ = \int \sec^{2} t dx \\ = tant + C \\ = \tan ({xe}^{x}) + C \\ Hence, the correct option is B. \end{array}$

Q.12

$Integrate the function given below . \frac{3 x^{2}}{x^{6} + 1}$

Ans.

$\begin{array}{l} Let t = x^{3} \Rightarrow \frac{dt}{dx} = 3 x^{2} \\ \Rightarrow dx = \frac{dt}{3 x^{2}} \\ ∴ \int \frac{3 x^{2}}{x^{6} + 1} dx = \int \frac{3 x^{2}}{{(x^{3})}^{2} + 1} dx \\ = \int \frac{3 x^{2}}{t^{2} + 1} . \frac{dt}{3 x^{2}} \\ = \int \frac{1}{t^{2} + 1} dt \\ = \tan^{- 1} t + C \\ = \tan^{- 1} (x^{3}) + C \end{array}$

Q.13

$Integrate the function given below . \frac{1}{\sqrt{1 + 4 x^{2}}}$

Ans.

$\begin{array}{l} We have, \frac{1}{\sqrt{1 + 4 x^{2}}} = \frac{1}{\sqrt{1 + {(2 x)}^{2}}} \\ Let t = 2 x \Rightarrow \frac{dt}{dx} = 2 \\ \Rightarrow dx = \frac{dt}{2} \\ ∴ \int \frac{1}{\sqrt{1 + 4 x^{2}}} dx = \int \frac{1}{\sqrt{1 + {(2 x)}^{2}}} dx \\ = \int \frac{1}{\sqrt{1 + t^{2}}} . \frac{dt}{2} \\ = \frac{1}{2} \int \frac{1}{\sqrt{1 + t^{2}}} dt \\ = \frac{1}{2} \log | t + \sqrt{t^{2} + 1} | + C \\ = \frac{1}{2} \log | 2 x + \sqrt{{(2 x)}^{2} + 1} | + C \\ = \frac{1}{2} \log | 2 x + \sqrt{4 x^{2} + 1} | + C \end{array}$

Q.14

$Integrate the function given below . \frac{1}{\sqrt{{(2 - x)}^{2} + 1}}$

Ans.

$\begin{array}{l} We have, \frac{1}{\sqrt{{(2 - x)}^{2} + 1}} \\ Let t = 2 - x \Rightarrow \frac{dt}{dx} = - 1 \\ \Rightarrow dx = - dt \\ ∴ \int \frac{1}{\sqrt{{(2 - x)}^{2} + 1}} dx = \int \frac{1}{\sqrt{t^{2} + 1}} (- dt) \\ = - \int \frac{1}{\sqrt{1 + t^{2}}} . dt \\ = - \log | t + \sqrt{t^{2} + 1} | + C [\begin{array}{l} ∵ \int \frac{1}{\sqrt{x^{2} + a^{2}}} dx \\ = \log | x + \sqrt{x^{2} + a^{2}} | \end{array}] \\ = - \log | (2 - x) + \sqrt{{(2 - x)}^{2} + 1} | + C \\ = \log | \frac{1}{(2 - x) + \sqrt{{(2 - x)}^{2} + 1}} | + C [∵ - logx = \log (\frac{1}{x})] \\ = \log | \frac{1}{(2 - x) + \sqrt{4 - 4 x + x^{2} + 1}} | + C \\ = \log | \frac{1}{(2 - x) + \sqrt{x^{2} - 4 x + 5}} | + C \end{array}$

Q.15

$Integrate the function given below . \frac{1}{\sqrt{9 - 25 x^{2}}}$

Ans.

$\begin{array}{l} We have, \frac{1}{\sqrt{9 - 25 x^{2}}} = \frac{1}{\sqrt{9 - {(5 x)}^{2}}} \\ Let t = 5 x \Rightarrow \frac{dt}{dx} = 5 \\ \Rightarrow dx = \frac{1}{5} dt \\ ∴ \int \frac{1}{\sqrt{9 - 25 x^{2}}} dx = \frac{1}{5} \int \frac{1}{\sqrt{9 - {(5 x)}^{2}}} dt \\ = \frac{1}{5} \int \frac{1}{\sqrt{3^{2} - t^{2}}} . dt \\ = \frac{1}{5} \sin^{- 1} (\frac{t}{3}) + C \\ = \frac{1}{5} \sin^{- 1} (\frac{5 x}{3}) + C \end{array}$

Q.16

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

Ans.

$\begin{array}{l} We have, \frac{3 x}{1 + 2 x^{4}} = \frac{3 x}{1 + {(\sqrt{2} x^{2})}^{2}} \\ Let t = \sqrt{2} x^{2} \Rightarrow \frac{dt}{dx} = 2 \sqrt{2} x \\ \Rightarrow dx = \frac{dt}{2 \sqrt{2} x} \\ ∴ \int \frac{3 x}{1 + 2 x^{4}} dx = \int \frac{3 x}{1 + {(\sqrt{2} x^{2})}^{2}} dx \\ = \int \frac{3 x}{1 + t^{2}} . \frac{dt}{2 \sqrt{2} x} \\ = \frac{3}{2 \sqrt{2}} \int \frac{1}{1 + t^{2}} dt \\ = \frac{3}{2 \sqrt{2}} \tan^{- 1} t + C = \frac{3}{2 \sqrt{2}} \tan^{- 1} (\sqrt{2} x^{2}) + C \end{array}$

Q.17

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

Ans.

$\begin{array}{l} We have, \frac{x^{2}}{1 - x^{6}} = \frac{x^{2}}{1 - {(x^{3})}^{2}} \\ Let t = x^{3} \Rightarrow \frac{dt}{dx} = 3 x^{2} \\ \Rightarrow dx = \frac{dt}{3 x^{2}} \\ ∴ \int \frac{x^{2}}{1 - x^{6}} dx = \int \frac{x^{2}}{1 - {(x^{3})}^{2}} dx \\ = \int \frac{x^{2}}{1 - t^{2}} . \frac{dt}{3 x^{2}} \\ = \frac{1}{3} \int \frac{1}{1 - t^{2}} dt \\ = \frac{1}{3} (\frac{1}{2} \log | \frac{1 + t}{1 - t} |) + C \\ = \frac{1}{6} \log | \frac{1 + x^{3}}{1 - x^{3}} | + C \end{array}$

Q.18

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

Ans.

$\begin{array}{l} We have, \frac{x - 1}{\sqrt{x^{2} - 1}} = \frac{x - 1}{\sqrt{x^{2} - 1}} \\ Let t = x^{2} - 1 \Rightarrow \frac{dt}{dx} = 2 x \\ \Rightarrow dx = \frac{dt}{2 x} \\ ∴ \int \frac{x - 1}{\sqrt{x^{2} - 1}} dx = \int \frac{x}{\sqrt{x^{2} - 1}} dx - \int \frac{1}{\sqrt{x^{2} - 1}} dx \\ = \int \frac{x}{\sqrt{t}} . \frac{dt}{2 x} - \int \frac{1}{\sqrt{x^{2} - 1}} dx \\ = \frac{1}{2} \int t^{- \frac{1}{2}} . dt - \log | x + \sqrt{x^{2} - 1} | + C \\ = \frac{1}{2} (2 \sqrt{t}) - \log | x + \sqrt{x^{2} - 1} | + C \\ = \sqrt{t} - \log | x + \sqrt{x^{2} - 1} | + C \\ = \sqrt{x^{2} - 1} - \log | x + \sqrt{x^{2} - 1} | + C \end{array}$

Q.19

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

Ans.

$\begin{array}{l} We have, \frac{x^{2}}{\sqrt{x^{6} + a^{6}}} = \frac{x^{2}}{\sqrt{a^{6} + {(x^{3})}^{2}}} \\ Let t = x^{3} \Rightarrow \frac{dt}{dx} = 3 x^{2} \\ \Rightarrow dx = \frac{dt}{3 x^{2}} \\ ∴ \int \frac{x^{2}}{\sqrt{x^{6} + a^{6}}} dx = \int \frac{x^{2}}{\sqrt{{(a^{3})}^{2} + {(x^{3})}^{2}}} dx \\ = \int \frac{x^{2}}{\sqrt{{(a^{3})}^{2} + t^{2}}} . \frac{dt}{3 x^{2}} \\ = \frac{1}{3} \int \frac{1}{\sqrt{{(a^{3})}^{2} + t^{2}}} dt \\ = \frac{1}{3} \log | t + \sqrt{t^{2} + {(a^{3})}^{2}} | + C \\ = \frac{1}{3} \log | x^{3} + \sqrt{{(x^{3})}^{2} + {(a^{3})}^{2}} | + C \\ = \frac{1}{3} \log | x^{3} + \sqrt{x^{6} + a^{6}} | + C \end{array}$

Q.20

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

Ans.

$\begin{array}{l} We have, \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 4}} = \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 2^{2}}} \\ Let t = tanx \Rightarrow \frac{dt}{dx} = \sec^{2} x \\ \Rightarrow dx = \frac{dt}{\sec^{2} x} \\ ∴ \int \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 4}} dx = \int \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 2^{2}}} dx \\ = \int \frac{\sec^{2} x}{\sqrt{t^{2} + 2^{2}}} . \frac{dt}{\sec^{2} x} \\ = \int \frac{1}{\sqrt{t^{2} + 2^{2}}} dt \\ = \log | t + \sqrt{t^{2} + {(2)}^{2}} | + C \\ = \log | tanx + \sqrt{\tan^{2} x + 4} | + C \end{array}$

Q.21

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

Ans.

$\begin{array}{l} We have, \\ \frac{1}{\sqrt{x^{2} + 2 x + 2}} = \frac{1}{\sqrt{{(x + 1)}^{2} + 1}} \\ Let t = x + 1 \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{x^{2} + 2 x + 2}} dx = \int \frac{1}{\sqrt{{(x + 1)}^{2} + 1}} dx \\ = \int \frac{1}{\sqrt{t^{2} + 1^{2}}} . dt \\ = \log | t + \sqrt{t^{2} + {(1)}^{2}} | + C \\ = \log | (x + 1) + \sqrt{{(x + 1)}^{2} + 1} | + C \\ = \log | (x + 1) + \sqrt{x^{2} + 2 x + 2} | + C \end{array}$

Q.22

$Integrate the function \frac{1}{9 x^{2} + 6 x + 5}$

Ans.

$\begin{array}{l} We have, \frac{1}{9 x^{2} + 6 x + 5} = \frac{1}{{(3 x + 1)}^{2} + 2^{2}} \\ Let t = 3 x + 1 \Rightarrow \frac{dt}{dx} = 3 \\ \Rightarrow dx = \frac{dt}{3} \\ ∴ \int \frac{1}{9 x^{2} + 6 x + 5} dx = \int \frac{1}{{(3 x + 1)}^{2} + 2^{2}} dx \\ = \int \frac{1}{t^{2} + 2^{2}} . \frac{dt}{3} \\ = \frac{1}{3} \int \frac{1}{t^{2} + 2^{2}} dt \\ = \frac{1}{3} (\frac{1}{2} \tan^{- 1} \frac{t}{2}) + C \\ = \frac{1}{6} \tan^{- 1} \frac{(3 x + 1)}{2} + C \end{array}$

Q.23

$Integrate the function \frac{1}{\sqrt{7 - 6 x - x^{2}}}$

Ans.

$\begin{array}{l} We have, \frac{1}{\sqrt{7 - 6 x - x^{2}}} = \frac{1}{\sqrt{16 - {(x + 3)}^{2}}} \\ Let t = x + 3 \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{7 - 6 x - x^{2}}} dx = \int \frac{1}{\sqrt{16 - {(x + 3)}^{2}}} dx \\ = \int \frac{1}{\sqrt{4^{2} - t^{2}}} dt \\ = \sin^{- 1} (\frac{t}{4}) + C \\ = \sin^{- 1} (\frac{x + 3}{4}) + C \end{array}$

Q.24

$Integrate the function \frac{1}{\sqrt{(x - 1) (x - 2)}}$

Ans.

$\begin{array}{l} We have, \frac{1}{\sqrt{(x - 1) (x - 2)}} = \frac{1}{\sqrt{x^{2} - 3 x + 2}} \\ = \frac{1}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}} \\ Let t = x - \frac{3}{2} \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{(x - 1) (x - 2)}} dx = \int \frac{1}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}} dx \\ = \int \frac{1}{\sqrt{t^{2} - {(\frac{1}{2})}^{2}}} dt \\ = \log | t + \sqrt{t^{2} - {(\frac{1}{2})}^{2}} | + C \\ = \log | (x - \frac{3}{2}) + \sqrt{{(x - \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}} | + C \\ = \log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + C \end{array}$

Q.25

$Integrate the function \frac{1}{\sqrt{8 + 3 x - x^{2}}}$

Ans.

$\begin{array}{l} We have, \frac{1}{\sqrt{8 + 3 x - x^{2}}} = \frac{1}{\sqrt{8 - (x^{2} - 3 x + \frac{9}{4} - \frac{9}{4})}} \\ = \frac{1}{\sqrt{8 + \frac{9}{4} - {(x - \frac{3}{2})}^{2}}} \\ = \frac{1}{\sqrt{\frac{41}{4} - {(x - \frac{3}{2})}^{2}}} \\ Let t = x - \frac{3}{2} \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{8 + 3 x - x^{2}}} dx = \int \frac{1}{\sqrt{\frac{41}{4} - {(x - \frac{3}{2})}^{2}}} dx \\ = \int \frac{1}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - t^{2}}} dt \\ = \sin^{- 1} (\frac{t}{\frac{\sqrt{41}}{2}}) + C \\ = \sin^{- 1} (\frac{2 t}{\sqrt{41}}) + C \\ = \sin^{- 1} {\frac{2 (x - \frac{3}{2})}{\sqrt{41}}} + C \\ = \sin^{- 1} {\frac{(2 x - 3)}{\sqrt{41}}} + C \end{array}$

Q.26

$Integrate the function \frac{1}{\sqrt{(x - a) (x - b)}}$

Ans.

$\begin{array}{l} We have, \frac{1}{\sqrt{(x - a) (x - b)}} = \frac{1}{\sqrt{x^{2} - (a + b) x + ab}} \\ = \frac{1}{\sqrt{{(x - \frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}}} \\ Let t = x - \frac{a + b}{2} \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{(x - a) (x - b)}} dx = \int \frac{1}{\sqrt{{(x - \frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}}} dx \\ = \int \frac{1}{\sqrt{t^{2} - {(\frac{a + b}{2})}^{2}}} dt \\ = \int \frac{1}{\sqrt{t^{2} - {(\frac{a + b}{2})}^{2}}} dt \\ = \log | t + \sqrt{t^{2} - {(\frac{a + b}{2})}^{2}} | + C \\ = \log | (x - \frac{a + b}{2}) + \sqrt{{(x - \frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}} | + C \\ = \log | (x - \frac{a + b}{2}) + \sqrt{x^{2} - (a + b) x + ab} | + C \end{array}$

Q.27

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

Ans.

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

Q.28

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

Ans.

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

Q.29

$Integrate the function \frac{5 x - 2}{1 + 2 x + 3 x^{2}}$

Ans.

$\begin{array}{l} We have, \frac{5 x - 2}{1 + 2 x + 3 x^{2}} \\ Here, 5 x - 2 = A \frac{d}{dx} (1 + 2 x + 3 x^{2}) + B \\ = A (6 x + 2) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 5 = 6 A and 2A + B = - 2 \\ \Rightarrow A = \frac{5}{6} and B = - 2 - 2 (\frac{5}{6}) = - 2 - \frac{5}{3} = - \frac{11}{3} \\ ∴ 5 x - 2 = \frac{5}{6} (6 x + 2) - \frac{11}{3} \\ ∴ \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} dx = \int \frac{\frac{5}{6} (6 x + 2) - \frac{11}{3}}{\sqrt{2 x^{2} + x - 3}} dx \\ = \frac{5}{6} \int \frac{(6 x + 2)}{\sqrt{1 + 2 x + 3 x^{2}}} dx - \frac{11}{3} \int \frac{1}{\sqrt{1 + 2 x + 3 x^{2}}} dx \\ Now, let t = 1 + 2 x + 3 x^{2} \Rightarrow \frac{dt}{dx} = 6 x + 2 \\ \Rightarrow dx = \frac{dt}{6 x + 2}, then \\ \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} dx = \frac{5}{6} \int \frac{(6 x + 2)}{1 + 2 x + 3 x^{2}} . \frac{dt}{6 x + 2} - \frac{11}{3} \int \frac{1}{1 + 2 x + 3 x^{2}} . dx \\ = \frac{5}{6} \int \frac{1}{t} . dt - \frac{11}{3} \int \frac{1}{3 {{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}}} . dx \\ = \frac{5}{6} \log | t | - \frac{11}{9} \int \frac{1}{{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} . dx \\ = \frac{5}{6} \log | 1 + 2 x + 3 x^{2} | - \frac{11}{9} \times \frac{1}{(\frac{\sqrt{2}}{3})} \tan^{- 1} (\frac{x + \frac{1}{3}}{\frac{\sqrt{2}}{3}}) + C \\ = \frac{5}{6} \log | 1 + 2 x + 3 x^{2} | - \frac{11}{3 \sqrt{2}} \tan^{- 1} (\frac{3 x + 1}{\sqrt{2}}) + C \end{array}$

Q.30

$Integrate the function \frac{6 x + 7}{\sqrt{(x - 5) (x - 4)}}$

Ans.

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

Q.31

$Integrate the function \frac{x + 2}{\sqrt{4 x - x^{2}}}$

Ans.

$\begin{array}{l} We have, \frac{x + 2}{\sqrt{4 x - x^{2}}} \\ Here, x + 2 = A \frac{d}{dx} (4 x - x^{2}) + B \\ x + 2 = A (4 - 2 x) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 1 = - 2 A and 4A + B = 2 \\ \Rightarrow A = - \frac{1}{2} and B = 2 - 4 (- \frac{1}{2}) = 4 \\ ∴ x + 2 = - \frac{1}{2} (4 - 2 x) + 4 \\ ∴ \int \frac{x + 2}{\sqrt{4 x - x^{2}}} dx = \int \frac{- \frac{1}{2} (4 - 2 x) + 4}{\sqrt{4 x - x^{2}}} dx \\ = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{4 x - x^{2}}} dx + \int \frac{4}{\sqrt{4 x - x^{2}}} dx \\ = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{4 x - x^{2}}} dx + \int \frac{4}{\sqrt{4 - (4 - 4 x + x^{2})}} . dx \\ = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{4 x - x^{2}}} dx + \int \frac{4}{\sqrt{2^{2} - {(2 - x)}^{2}}} dx \\ Now, let t = 4 x - x^{2} \Rightarrow \frac{dt}{dx} = 4 - 2 x \\ \Rightarrow dx = \frac{dt}{4 - 2 x}, then \\ \int \frac{x + 2}{\sqrt{4 x - x^{2}}} dx = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{t}} \frac{dt}{4 - 2 x} + \int \frac{4}{\sqrt{2^{2} - {(2 - x)}^{2}}} dx \\ = - \frac{1}{2} \times 2 \sqrt{t} - 4 \sin^{- 1} (\frac{2 - x}{2}) + C \\ = - \sqrt{4 x - x^{2}} - 4 \sin^{- 1} (\frac{2 - x}{2}) + C \end{array}$

Q.32

$Integrate the function \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}}$

Ans.

$\begin{array}{l} We have, \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} \\ \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} dx = \frac{1}{2} \int \frac{2 (x + 2)}{\sqrt{x^{2} + 2 x + 3}} dx \\ = \frac{1}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 2 x + 3}} dx \\ = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} dx + \frac{1}{2} \int \frac{2}{\sqrt{x^{2} + 2 x + 3}} dx \\ Let t = x^{2} + 2 x + 3 \Rightarrow \frac{dt}{dx} = 2 x + 2 \\ \Rightarrow dx = \frac{dt}{2 x + 2}, then \\ \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} dx = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{t}} \frac{dt}{2 x + 2} + \frac{1}{2} \int \frac{2}{\sqrt{x^{2} + 2 x + 3}} dx \\ = \frac{1}{2} \int \frac{1}{\sqrt{t}} dt + \int \frac{1}{\sqrt{{(x + 1)}^{2} + {(\sqrt{2})}^{2}}} dx \\ = \frac{1}{2} . 2 \sqrt{t} + \log | (x + 1) + \sqrt{{(x + 1)}^{2} + {(\sqrt{2})}^{2}} | + C \\ = \sqrt{x^{2} + 2 x + 3} + \log | (x + 1) + \sqrt{x^{2} + 2 x + 3} | + C \end{array}$

Q.33

$Integrate the function \frac{x + 3}{x^{2} - 2 x - 5}$

Ans.

$\begin{array}{l} We have, \frac{x + 3}{x^{2} - 2 x - 5} \\ Here, x + 3 = A \frac{d}{dx} (x^{2} - 2 x - 5) + B \\ x + 3 = A (2 x - 2) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 1 = 2 A and - 2A + B = 3 \\ \Rightarrow A = \frac{1}{2} and B = 3 + 2 (\frac{1}{2}) = 4 \\ ∴ x + 3 = \frac{1}{2} (2 x - 2) + 4 \\ \Rightarrow \int \frac{x + 3}{x^{2} - 2 x - 5} dx = \int \frac{\frac{1}{2} (2 x - 2) + 4}{x^{2} - 2 x - 5} dx \\ = \frac{1}{2} \int \frac{(2 x - 2)}{x^{2} - 2 x - 5} dx + 4 \int \frac{1}{x^{2} - 2 x - 5} dx \\ Let t = x^{2} - 2 x - 5 \Rightarrow \frac{dt}{dx} = 2 x - 2 \\ \Rightarrow dx = \frac{dt}{2 x - 2}, then \\ \Rightarrow \int \frac{x + 3}{x^{2} - 2 x - 5} dx = \frac{1}{2} \int \frac{(2 x - 2)}{t} \frac{dt}{2 x - 2} + 4 \int \frac{1}{{(x - 1)}^{2} - 6} dx \\ = \frac{1}{2} \int \frac{1}{t} dt + 4 \int \frac{1}{{(x - 1)}^{2} - {(\sqrt{6})}^{2}} dx \\ = \frac{1}{2} \log | t | + 4 \times \frac{1}{2 \sqrt{6}} \log (\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}) + C \\ = \frac{1}{2} \log | x^{2} - 2 x - 5 | + \frac{2}{\sqrt{6}} \log (\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}) + C \end{array}$

Q.34

$Integrate the function \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}}$

Ans.

$\begin{array}{l} We have, \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} \\ Here, 5 x + 3 = A \frac{d}{dx} (x^{2} + 4 x + 10) + B \\ 5 x + 3 = A (2 x + 4) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 5 = 2 A and 4A + B = 3 \\ \Rightarrow A = \frac{5}{2} and B = 3 - 4 (\frac{5}{2}) = - 7 \\ ∴ 5 x + 3 = \frac{5}{2} (2 x + 4) - 7 \\ ∴ \int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} dx = \int \frac{\frac{5}{2} (2 x + 4) - 7}{\sqrt{x^{2} + 4 x + 10}} dx \\ = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{x^{2} + 4 x + 10}} dx - \int \frac{7}{\sqrt{x^{2} + 4 x + 10}} dx \\ = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{x^{2} + 4 x + 10}} dx - \int \frac{7}{\sqrt{{(x + 2)}^{2} + 6}} dx \\ = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{x^{2} + 4 x + 10}} dx - 7 \int \frac{1}{\sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}}} dx \\ Now, let t = x^{2} + 4 x + 10 \Rightarrow \frac{dt}{dx} = 2 x + 4 \\ \Rightarrow dx = \frac{dt}{2 x + 4}, then \\ \int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} dx = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{t}} \frac{dt}{2 x + 4} - 7 \int \frac{1}{\sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}}} dx \\ = \frac{5}{2} \times 2 \sqrt{t} - 7 \log | (x + 2) + \sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}} | + C \\ = 5 \sqrt{x^{2} + 4 x + 10} - 7 \log | (x + 2) + \sqrt{x^{2} + 4 x + 10} | + C \end{array}$

Q.35

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

Ans.

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

Q.36

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

Ans.

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

Q.37

$Integrate the rational function \frac{x}{(x + 1) (x + 2)}$

Ans.

$\begin{array}{l} Let \frac{x}{(x + 1) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} \\ x = A (x + 2) + B (x + 1) \\ x = (A + B) x + 2 A + B \\ \Rightarrow A + B = 1 and 2 A + B = 0 \\ \Rightarrow A = - 1 and B = 2 \\ So, \frac{x}{(x + 1) (x + 2)} = \frac{- 1}{x + 1} + \frac{2}{x + 2} \\ ∴ \int \frac{x}{(x + 1) (x + 2)} dx = \int \frac{- 1}{x + 1} dx + \int \frac{2}{x + 2} dx \\ = \log {(x + 1)}^{- 1} + \log {(x + 2)}^{2} + C \\ = \log \frac{{(x + 2)}^{2}}{(x + 1)} + C \end{array}$

Q.38

$Integrate the rational function \frac{1}{x^{2} - 9}$

Ans.

$\begin{array}{l} Let \frac{1}{x^{2} - 9} = \frac{1}{(x + 3) (x - 3)} \\ = \frac{A}{x + 3} + \frac{B}{x - 3} \\ 1 = A (x - 3) + B (x + 3) \\ 1 = (A + B) x + 3 (- A + B) \\ \Rightarrow A + B = 0 and 3 (- A + B) = 1 \\ \Rightarrow A = - \frac{1}{6} and B = \frac{1}{6} \\ So, \frac{1}{(x + 3) (x - 3)} = \frac{- 1}{6 (x + 3)} + \frac{1}{6 (x - 3)} \\ ∴ \int \frac{1}{(x + 3) (x - 3)} dx = \int \frac{- 1}{6 (x + 3)} dx + \int \frac{1}{6 (x - 3)} dx \\ = - \frac{1}{6} \log | x + 3 | + \frac{1}{6} \log | x - 3 | + C \\ = \frac{1}{6} \log | \frac{x - 3}{x + 3} | + C \end{array}$

Q.39

$Integrate the rational function \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)}$

Ans.

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

Q.40

$Integrate the rational function \frac{x}{(x - 1) (x - 2) (x - 3)}$

Ans.

$\begin{array}{l} Let \frac{x}{(x - 1) (x - 2) (x - 3)} \\ = \frac{A}{(x - 1)} + \frac{B}{(x - 2)} + \frac{C}{(x - 3)} \\ x = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) . .. (i) \\ Substituting x = 1, 2 and 3 respectively in equation (i), we get \\ A = \frac{1}{2}, B = - 2 and C = \frac{3}{2} \\ \Rightarrow \frac{x}{(x - 1) (x - 2) (x - 3)} = \frac{1}{2 (x - 1)} + \frac{- 2}{x - 2} + \frac{3}{2 (x - 3)} \\ ∴ \int \frac{x}{(x - 1) (x - 2) (x - 3)} dx = \frac{1}{2} \int \frac{1}{x - 1} dx - 2 \int \frac{1}{x - 2} dx + \frac{3}{2} \int \frac{1}{(x - 3)} dx \\ = \frac{1}{2} \log | x - 1 | - 2 \log | x - 2 | + \frac{3}{2} \log | x - 3 | + C \end{array}$

Q.41

$Integrate the rational function \frac{2 x}{x^{2} + 3 x + 2}$

Ans.

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

Q.42

$Integrate the rational function \frac{1 - x^{2}}{x (1 - 2 x)}$

Ans.

$\begin{array}{l} \frac{1 - x^{2}}{x (1 - 2 x)} = \frac{1}{2} + \frac{1}{2} \{\frac{2 - x}{x (1 - 2 x)}\} . .. (i) \\ Let \frac{2 - x}{x (1 - 2 x)} = \frac{A}{x} + \frac{B}{1 - 2 x} \\ \Rightarrow 2 - x = A (1 - 2 x) + Bx \\ Substituting x = 0 and \frac{1}{2} respectively in equation (i), we get \\ A = 2, B = 3 \\ ∴ \frac{2 - x}{x (1 - 2 x)} = \frac{2}{x} + \frac{3}{1 - 2 x} \\ From equation (i), we have \\ \int \frac{1 - x^{2}}{x (1 - 2 x)} dx = \int \frac{1}{2} dx + \frac{1}{2} \int \frac{2}{x} dx + \frac{1}{2} \int \frac{3}{1 - 2 x} dx \\ = \frac{1}{2} x + \frac{1}{2} (2 \log |x| - \frac{3}{4} \log |1 - 2 x|) + C \\ = \frac{1}{2} x + \log |x| - \frac{3}{4} \log |1 - 2 x| + C \end{array}$

Q.43

$Integrate the rational function \frac{x}{(x^{2} + 1) (x - 1)}$

Ans.

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

Q.44

$Integrate the rational function \frac{x}{{(x - 1)}^{2} (x + 2)}$

Ans.

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

Q.45

$Integrate the rational function \frac{3 x + 5}{x^{3} - x^{2} - x + 1}$

Ans.

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

Q.46

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

Ans.

$\begin{array}{l} \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} = \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} \\ = \frac{A}{(x + 1)} + \frac{B}{(x - 1)} + \frac{C}{(2 x + 3)} \\ 2 x - 3 = A (x - 1) (2 x + 3) + B (x - 1) (2 x + 3) + C (x - 1) (x + 1) . .. (i) \\ Substituting x = - 1, 1 and - \frac{3}{2} respectively in equation (i), we get \\ A = \frac{5}{2}, B = - \frac{1}{10} and C = - \frac{24}{5} \\ \Rightarrow \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{5}{2 (x + 1)} - \frac{1}{10 (x - 1)} - \frac{24}{5 (2 x + 3)} \\ ∴ \int \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} dx = \frac{5}{2} \int \frac{1}{(x + 1)} dx - \frac{1}{10} \int \frac{1}{(x - 1)} dx \\ - \frac{24}{5} \int \frac{1}{(2 x + 3)} dx \\ = \frac{5}{2} \log | (x + 1) | - \frac{1}{10} \log | (x - 1) | \\ - \frac{24}{5 \times 2} \log | (2 x + 3) | + D \\ = \frac{5}{2} \log | (x + 1) | - \frac{1}{10} \log | (x - 1) | \\ - \frac{12}{5} \log | (2 x + 3) | + D \end{array}$

Q.47

$Integrate the rational function \frac{5 x}{(x + 1) (x^{2} - 4)}$

Ans.

$\begin{array}{l} \frac{5 x}{(x + 1) (x^{2} - 4)} = \frac{5 x}{(x + 1) (x - 2) (x + 2)} \\ = \frac{A}{(x + 1)} + \frac{B}{(x + 2)} + \frac{C}{(x - 2)} \\ 5 x = A (x - 2) (x + 2) + B (x + 1) (x - 2) + C (x + 1) (x + 2) . .. (i) \\ Substituting x = - 1, - 2 and 2 respectively in equation (i), we get \\ A = \frac{5}{3}, B = - \frac{5}{2} and C = \frac{5}{6} \\ \Rightarrow \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{5}{3 (x + 1)} - \frac{5}{2 (x + 2)} + \frac{5}{6 (x - 2)} \\ ∴ \int \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} dx = \frac{5}{3} \int \frac{1}{(x + 1)} dx - \frac{5}{2} \int \frac{1}{(x + 2)} dx \\ + \frac{5}{6} \int \frac{1}{(x - 2)} dx \\ = \frac{5}{3} \log |(x + 1)| - \frac{5}{2} \log |(x + 2)| \\ + \frac{5}{6} \log |(x - 2)| + D \end{array}$

Q.48

$Integrate the rational function \frac{x^{3} + x + 1}{x^{2} - 1}$

Ans.

$\begin{array}{l} ∵ \frac{x^{3} + x + 1}{x^{2} - 1} = x + \frac{2 x + 1}{x^{2} - 1} \\ Let, \frac{2 x + 1}{x^{2} - 1} = \frac{A}{x + 1} + \frac{B}{x - 1} \\ 2 x + 1 = A (x - 1) + B (x + 1) . .. (i) \\ Substituting x = - 1 and 1 respectively in equation (i), we get \\ 2 (- 1) + 1 = A (- 1 - 1) + B (- 1 + 1) \\ - 1 = A (- 2) \Rightarrow A = \frac{1}{2} \\ and B = \frac{3}{2} \\ \Rightarrow \frac{x^{3} + x + 1}{x^{2} - 1} = x + \frac{1}{2 (x + 1)} + \frac{3}{2 (x - 1)} \\ \int \frac{x^{3} + x + 1}{x^{2} - 1} dx = \int x dx + \frac{1}{2} \int \frac{1}{(x + 1)} dx + \frac{3}{2} \int \frac{1}{x - 1} dx \\ = \frac{x^{2}}{2} + \frac{1}{2} \log | x + 1 | + \frac{3}{2} \log | x - 1 | + C \end{array}$

Q.49

$Integrate the rational function \frac{2}{(1 - x) (1 + x^{2})}$

Ans.

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

Q.50

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

Ans.

$\begin{array}{l} We have, \frac{3 x - 1}{{(x + 2)}^{2}} \\ Let \frac{3 x - 1}{{(x + 2)}^{2}} = \frac{A}{(x + 2)} + \frac{B}{{(x + 2)}^{2}} \\ 3 x - 1 = A (x + 2) + B \\ Equating {the coefficients of x}^{2}, x and constant term, we get \\ A = 3 \\ 2 A + B = - 1 \\ Solving, the above equations, we get \\ A = 3 and B = - 7 \\ ∴ \frac{3 x - 1}{{(x + 2)}^{2}} = \frac{3}{(x + 2)} - \frac{7}{{(x + 2)}^{2}} \\ \Rightarrow \int \frac{3 x - 1}{{(x + 2)}^{2}} dx = \int \frac{3}{(x + 2)} dx - \int \frac{7}{{(x + 2)}^{2}} dx \\ = 3 \log | x + 2 | - 7 (\frac{- 1}{x + 2}) + C \\ = 3 \log | x + 2 | + \frac{7}{x + 2} + C \end{array}$

Q.51

$Integrate the rational function \frac{1}{x^{4} - 1}$

Ans.

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

Q.52

$Integrate the rational function \frac{1}{x (x^{n} - 1)}$

Ans.

$\begin{array}{l} We have, \frac{1}{x (x^{n} - 1)} = \frac{1}{x (x^{n} - 1)} \times \frac{x^{n - 1}}{x^{n - 1}} \\ = \frac{x^{n - 1}}{x^{n} (x^{n} - 1)} \\ Let t = x^{n} \Rightarrow \frac{dt}{dx} = {nx}^{n - 1} \Rightarrow \frac{dt}{{nx}^{n - 1}} = dx \\ ∴ \int \frac{1}{x (x^{n} - 1)} dx = \int \frac{x^{n - 1}}{x^{n} (x^{n} - 1)} dx \\ = \frac{1}{n} \int \frac{1}{t (t - 1)} dt \\ Let \frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \\ 1 = A (t - 1) + Bt \\ = t (A + B) - A \\ Substituting t = 0 and 1 respectively, we get \\ A = 1 and B = - 1 \\ ∴ \frac{1}{t (t - 1)} = \frac{1}{t} + \frac{- 1}{t - 1} \\ \Rightarrow \int \frac{1}{x (x^{n} - 1)} dx = \frac{1}{n} \int (\frac{1}{t} - \frac{1}{t - 1}) dt \\ = \frac{1}{n} \int \frac{1}{t} dt - \frac{1}{n} \int \frac{1}{t - 1} dt \\ = \frac{1}{n} \log | t | - \frac{1}{n} \log | t - 1 | + C \\ = \frac{1}{n} \log | x^{n} | - \frac{1}{n} \log | x^{n} - 1 | + C \\ = \frac{1}{n} \log | \frac{x^{n}}{x^{n} - 1} | + C \end{array}$

Q.53

$Integrate the rational function \frac{cosx}{(1 - sinx) (2 - sinx)}$

Ans.

$\begin{array}{l} We have, \frac{cosx}{(1 - sinx) (2 - sinx)} \\ Let t = sinx \Rightarrow \frac{dt}{dx} = cosx \Rightarrow \frac{dt}{cosx} = dx \\ ∴ \int \frac{cosx}{(1 - sinx) (2 - sinx)} dx = \int \frac{cosx}{(1 - t) (2 - t)} \frac{dt}{cosx} \\ = \int \frac{1}{(1 - t) (2 - t)} dt \\ Let \frac{1}{(1 - t) (2 - t)} = \frac{A}{1 - t} + \frac{B}{2 - t} \\ 1 = A (2 - t) + B (1 - t) . .. (1) \\ Substituting t = 1 and 2 respectively in eqution (1), we get \\ A = 1 and B = - 1 \\ ∴ \frac{1}{(1 - t) (2 - t)} = \frac{1}{1 - t} - \frac{1}{2 - t} \\ So, \\ \int \frac{cosx}{(1 - sinx) (2 - sinx)} dx = \int \frac{1}{(1 - t) (2 - t)} dt \\ = \int \frac{1}{1 - t} dt - \int \frac{1}{2 - t} dt \\ = - \log | 1 - t | + \log | 2 - t | + C \\ = \log | \frac{2 - t}{1 - t} | + C \\ = \log | \frac{2 - sinx}{1 - sinx} | + C \end{array}$

Q.54

$Integrate the rational function \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)}$

Ans.

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

Q.55

$Integrate the rational function \frac{2 x}{(x^{2} + 1) (x^{2} + 3)}$

Ans.

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

Q.56

$Integrate the rational function \frac{1}{x (x^{4} - 1)}$

Ans.

$\begin{array}{l} We have, \frac{1}{x (x^{4} - 1)} = \frac{1}{x (x^{4} - 1)} \times \frac{x^{3}}{x^{3}} \\ = \frac{x^{3}}{x^{4} (x^{4} - 1)} \\ Let t = x^{4} \Rightarrow \frac{dt}{dx} = 4 x^{4 - 1} \Rightarrow \frac{dt}{4 x^{3}} = dx \\ ∴ \int \frac{1}{x (x^{4} - 1)} dx = \int \frac{x^{3}}{x^{4} (x^{4} - 1)} dx \\ = \int \frac{x^{3}}{t (t - 1)} \frac{dt}{4 x^{3}} \\ = \frac{1}{4} \int \frac{1}{t (t - 1)} dt \\ Let \frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \\ 1 = A (t - 1) + Bt \\ Substituting t = 0 and 1 respectively, we get \\ A = - 1 and B = 1 \\ ∴ \frac{1}{t (t - 1)} = \frac{- 1}{t} + \frac{1}{t - 1} \\ \Rightarrow \int \frac{1}{x (x^{4} - 1)} dx = \frac{1}{4} \int (- \frac{1}{t} + \frac{1}{t - 1}) dt \\ = - \frac{1}{4} \int \frac{1}{t} dt + \frac{1}{4} \int \frac{1}{t - 1} dt \\ = - \frac{1}{4} \log | t | + \frac{1}{4} \log | t - 1 | + C \\ = - \frac{1}{4} \log | x^{4} | + \frac{1}{4} \log | x^{4} - 1 | + C \\ = \frac{1}{4} \log | \frac{x^{4} - 1}{x^{4}} | + C \end{array}$

Q.57

$Integrate the rational function \frac{1}{(e^{x} - 1)}$

Ans.

$\begin{array}{l} \int \frac{1}{(e^{x} - 1)} dx \\ Let t = e^{x} \Rightarrow \frac{dt}{dx} = e^{x} \\ dx = \frac{dt}{e^{x}} = \frac{dt}{t} \\ ∴ \int \frac{1}{(e^{x} - 1)} dx = \int \frac{1}{(t - 1)} \frac{dt}{t} \\ = \int \frac{1}{t (t - 1)} dt \\ Let \frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \\ = \frac{A (t - 1) + Bt}{t (t - 1)} \\ \Rightarrow 1 = A (t - 1) + Bt . .. (i) \\ Putting t = 0 and 1 respectively in equation (i), we get \\ A = - 1 and B = 1 \\ ∴ \frac{1}{t (t - 1)} = \frac{- 1}{t} + \frac{1}{t - 1} \\ \Rightarrow \int \frac{1}{t (t - 1)} dt = \int \frac{- 1}{t} dt + \int \frac{1}{t - 1} dt \\ = - \log | t | + \log | t - 1 | + C \\ = \log | \frac{t - 1}{t} | + C \\ ∴ \int \frac{1}{(e^{x} - 1)} dx = \log | \frac{e^{x} - 1}{e^{x}} | + C \end{array}$

Q.58

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

Ans.

$\begin{array}{l} We have, \int \frac{xdx}{(x - 1) (x - 2)} dx \\ Let \frac{x}{(x - 1) (x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \\ = \frac{A (x - 2) + B (x - 1)}{(x - 1) (x - 2)} \\ \Rightarrow x = A (x - 2) + B (x - 1) . .. (i) \\ Substituting x = 1 and 2, in equation (i), we get \\ A = - 1 and B = 2 \\ \Rightarrow \frac{x}{(x - 1) (x - 2)} = \frac{- 1}{x - 1} + \frac{2}{x - 2} \\ ∴ \int \frac{x}{(x - 1) (x - 2)} dx = \int \frac{- 1}{x - 1} dx + \int \frac{2}{x - 2} dx \\ = - log | x - 1 | + 2 \log | x - 2 | + C \\ = \log | \frac{{(x - 2)}^{2}}{(x - 1)} | + C \\ Hence, the option (B) is correct. \end{array}$

Q.59

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

Ans.

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

Q.60

Integrate the function x sin x.

Ans.

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

Q.61

Integrate the functions
x sin 3x

Ans.

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

Q.62

Integrate the functions
x² e^x

Ans.

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

Q.63

Integrate the functions
x logx

Ans.

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

Q.64

Integrate the functions
x log 2x

Ans.

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

Q.65

Integrate the functions
x sin^-1x

Ans.

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

Q.66

Integrate the functions
x tan^-1x

Ans.

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

Q.67

Integrate the functions
x cos^-1x

Ans.

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

Q.68

Integrate the functions
(sin^–1x)²

Ans.

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

Q.69

Integrate the functions
x² logx

Ans.

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

Q.70

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

Ans.

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

Q.71

Integrate the functions x sec²x

Ans.

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

Q.72

Integrate the functions tan^-1x

Ans.

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

Q.73

Integrate the functions x (logx)²

Ans.

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

Q.74

Integrate the functions (x²+1) logx

Ans.

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

Q.75

Integrate the functions e^x(sinx + cosx)

Ans.

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

Q.76

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

Ans.

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

Q.77

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

Ans.

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

Q.78

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

Ans.

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

Q.79

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

Ans.

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

Q.80

Integrate the functions e^2x sinx

Ans.

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

Q.81

$\begin{array}{l} Integrate the functions \\ \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \end{array}$

Ans.

$\begin{array}{l} We have, \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \\ Let I = \int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx \\ Let x = tanθ \Rightarrow \frac{dx}{dθ} = \sec^{2} θ \\ ∴ I = \int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx \\ = \int \sin^{- 1} (\frac{2 tanθ}{1 + \tan^{2} θ}) \sec^{2} θ dθ \\ = \int \sin^{- 1} (\sin 2 θ) \sec^{2} θ dθ \\ = \int 2 θ \sec^{2} θ dθ \\ = 2 [θ \int \sec^{2} θ dθ - (\int \frac{d}{dθ} θ \int \sec^{2} θ dθ) dθ] \\ = 2 [θtanθ - \int 1 tanθ dθ] \\ = 2 θtanθ - 2 \log | secθ | + C \\ = 2 θtanθ - 2 \log | \sqrt{1 + \tan^{2} θ} | + C \\ I = 2 {xtan}^{- 1} x - 2 \log | \sqrt{1 + x^{2}} | + C . \end{array}$

Q.82

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

Ans.

$\begin{array}{l} We have \int x^{2} e^{x^{3}} dx \\ Let t = x^{3} \Rightarrow \frac{dt}{dx} = 3 x^{2} \\ ∴ \int x^{2} e^{x^{3}} dx = \int x^{2} e^{t} \frac{dt}{3 x^{2}} \\ = \frac{1}{3} \int e^{t} dt \\ = \frac{1}{3} e^{x^{3}} + C \\ Hence, the correct option is A. \end{array}$

Q.83

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

Ans.

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

Q.84

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

Ans.

$\begin{array}{l} We have, \sqrt{4 - x^{2}} \\ Let I = \int \sqrt{4 - x^{2}} dx \\ = \int \sqrt{2^{2} - x^{2}} dx \\ [∵ \int \sqrt{a^{2} - x^{2}} dx = \frac{1}{2} x \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C] \\ = \frac{1}{2} x \sqrt{2^{2} - x^{2}} + \frac{2^{2}}{2} \sin^{- 1} \frac{x}{2} + C \\ = \frac{x}{2} \sqrt{4 - x^{2}} + 2 \sin^{- 1} \frac{x}{2} + C \end{array}$

Q.85

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

Ans.

$\begin{array}{l} We have, \sqrt{1 - 4 x^{2}} \\ Let I = \int \sqrt{1 - 4 x^{2}} dx \\ = \int \sqrt{1^{2} - {(2 x)}^{2}} dx \\ [Let t = 2 x \Rightarrow \frac{dt}{dx} = 2 \Rightarrow dx = \frac{dt}{2}] \\ = \int \sqrt{1^{2} - t^{2}} \frac{dt}{2} \\ [∵ \int \sqrt{a^{2} - x^{2}} dx = \frac{1}{2} x \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C] \\ = \frac{1}{2} (\frac{1}{2} t \sqrt{1^{2} - t^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{t}{1}) + C \\ = \frac{1}{2} (\frac{2 x}{2} \sqrt{1 - 4 x^{2}} + \frac{1}{2} \sin^{- 1} \frac{2 x}{1}) + C \\ = \frac{x}{2} \sqrt{1 - 4 x^{2}} + \frac{1}{4} \sin^{- 1} 2 x + C \end{array}$

Q.86

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

Ans.

$\begin{array}{l} We have, \sqrt{x^{2} + 4 x + 6} = \sqrt{{(x + 2)}^{2} + 2} \\ Let I = \int \sqrt{{(x + 2)}^{2} + {(\sqrt{2})}^{2}} dx \\ = \int \sqrt{t^{2} + {(\sqrt{2})}^{2}} dt [Let t = x + 2 \Rightarrow \frac{dt}{dx} = 1] \\ [∵ \int \sqrt{x^{2} + a^{2}} dx = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} |] \\ = \frac{t}{2} \sqrt{t^{2} + {(\sqrt{2})}^{2}} + \frac{{(\sqrt{2})}^{2}}{2} \log | t + \sqrt{t^{2} + {(\sqrt{2})}^{2}} | + C \\ Putting t = x + 2, we get \\ = \frac{(x + 2)}{2} \sqrt{{(x + 2)}^{2} + {(\sqrt{2})}^{2}} + \frac{2}{2} \log | (x + 2) + \sqrt{{(x + 2)}^{2} + {(\sqrt{2})}^{2}} | + C \\ = \frac{(x + 2)}{2} \sqrt{x^{2} + 4 x + 6} + \log | (x + 2) + \sqrt{x^{2} + 4 x + 6} | + C \end{array}$

Q.87

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

Ans.

$\begin{array}{l} We have, \sqrt{x^{2} + 4 x + 1} = \sqrt{{(x + 2)}^{2} - 3} \\ Let I = \int \sqrt{{(x + 2)}^{2} - {(\sqrt{3})}^{2}} dx \\ = \int \sqrt{t^{2} - {(\sqrt{3})}^{2}} dt [Let t = x + 2 \Rightarrow \frac{dt}{dx} = 1] \\ [∵ \int \sqrt{x^{2} - a^{2}} dx = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] \\ = \frac{t}{2} \sqrt{t^{2} - {(\sqrt{3})}^{2}} - \frac{{(\sqrt{3})}^{2}}{2} \log | t + \sqrt{t^{2} - {(\sqrt{3})}^{2}} | + C \\ Putting t = x + 2, we get \\ = \frac{(x + 2)}{2} \sqrt{{(x + 2)}^{2} - {(\sqrt{3})}^{2}} - \frac{3}{2} \log | (x + 2) + \sqrt{{(x + 2)}^{2} - {(\sqrt{3})}^{2}} | + C \\ = \frac{(x + 2)}{2} \sqrt{x^{2} + 4 x + 1} - \frac{3}{2} \log | (x + 2) + \sqrt{x^{2} + 4 x + 1} | + C \end{array}$

Q.88

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

Ans.

$\begin{array}{l} We have, \sqrt{1 - 4 x - x^{2}} = \sqrt{5 - (x^{2} + 4 x + 4)} \\ = \sqrt{{(\sqrt{5})}^{2} - {(x + 2)}^{2}} \\ Let I = \int \sqrt{{(\sqrt{5})}^{2} - {(x + 2)}^{2}} dx \\ = \int \sqrt{{(\sqrt{5})}^{2} - t^{2}} dt [Let t = x + 2 \Rightarrow \frac{dt}{dx} = 1] \\ [∵ \int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a}] \\ = \frac{t}{2} \sqrt{{(\sqrt{5})}^{2} - t^{2}} + \frac{{(\sqrt{5})}^{2}}{2} \sin^{- 1} \frac{t}{(\sqrt{5})} + C \\ Putting t = x + 2, we get \\ = \frac{(x + 2)}{2} \sqrt{{(\sqrt{5})}^{2} - {(x + 2)}^{2}} + \frac{{(\sqrt{5})}^{2}}{2} \sin^{- 1} \frac{(x + 2)}{(\sqrt{5})} + C \\ = \frac{(x + 2)}{2} \sqrt{1 - 4 x - x^{2}} + \frac{5}{2} \sin^{- 1} \frac{(x + 2)}{(\sqrt{5})} + C \end{array}$

Q.89

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

Ans.

$\begin{array}{l} We have, \sqrt{x^{2} + 4 x - 5} = \sqrt{{(x + 2)}^{2} - 9} \\ Let I = \int \sqrt{{(x + 2)}^{2} - {(3)}^{2}} dx \\ = \int \sqrt{t^{2} - {(3)}^{2}} dt [Let t = x + 2 \Rightarrow \frac{dt}{dx} = 1] \\ [∵ \int \sqrt{x^{2} - a^{2}} dx = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] \\ = \frac{t}{2} \sqrt{t^{2} - {(3)}^{2}} - \frac{{(3)}^{2}}{2} \log | t + \sqrt{t^{2} - {(3)}^{2}} | + C \\ Putting t = x + 2, we get \\ = \frac{(x + 2)}{2} \sqrt{{(x + 2)}^{2} - {(3)}^{2}} - \frac{9}{2} \log | (x + 2) + \sqrt{{(x + 2)}^{2} - {(3)}^{2}} | + C \\ = \frac{(x + 2)}{2} \sqrt{x^{2} + 4 x - 5} - \frac{9}{2} \log | (x + 2) + \sqrt{x^{2} + 4 x - 5} | + C \end{array}$

Q.90

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

Ans.

$\begin{array}{l} We have, \sqrt{1 + 3 x - x^{2}} = \sqrt{1 - (x^{2} - 3 x + \frac{9}{4} - \frac{9}{4})} \\ = \sqrt{{(\frac{\sqrt{13}}{2})}^{2} - {(x - \frac{3}{2})}^{2}} \\ Let I = \int \sqrt{{(\frac{\sqrt{13}}{2})}^{2} - {(x - \frac{3}{2})}^{2}} dx \\ = \int \sqrt{{(\frac{\sqrt{13}}{2})}^{2} - t^{2}} dt [Let t = x - \frac{3}{2} \Rightarrow \frac{dt}{dx} = 1] \\ [∵ \int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a}] \\ = \frac{t}{2} \sqrt{{(\frac{\sqrt{13}}{2})}^{2} - t^{2}} + \frac{{(\frac{\sqrt{13}}{2})}^{2}}{2} \sin^{- 1} \frac{t}{(\frac{\sqrt{13}}{2})} + C \\ Putting t = x - \frac{3}{2}, we get \\ = \frac{(x - \frac{3}{2})}{2} \sqrt{{(\frac{\sqrt{13}}{2})}^{2} - {(x - \frac{3}{2})}^{2}} + \frac{{(\frac{\sqrt{13}}{2})}^{2}}{2} \sin^{- 1} \frac{(x - \frac{3}{2})}{(\frac{\sqrt{13}}{2})} + C \\ = \frac{(2 x - 3)}{4} \sqrt{1 + 3 x - x^{2}} + \frac{13}{8} \sin^{- 1} \frac{(2 x - 3)}{(\sqrt{13})} + C \end{array}$

Q.91

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

Ans.

$\begin{array}{l} We have, \sqrt{x^{2} + 3 x} = \sqrt{(x^{2} + 3 x + \frac{9}{4}) - \frac{9}{4}} \\ = \sqrt{{(x + \frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}} \\ Let I = \int \sqrt{{(x + \frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}} dx \\ = \int \sqrt{t^{2} - {(\frac{3}{2})}^{2}} dt [Let t = x + \frac{3}{2} \Rightarrow \frac{dt}{dx} = 1] \\ [∵ \int \sqrt{x^{2} - a^{2}} dx = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] \\ = \frac{t}{2} \sqrt{t^{2} - {(\frac{3}{2})}^{2}} - \frac{{(\frac{3}{2})}^{2}}{2} \log | t + \sqrt{t^{2} - {(\frac{3}{2})}^{2}} | + C \\ Putting t = x + \frac{3}{2}, we get \\ = \frac{(x + \frac{3}{2})}{2} \sqrt{{(x + \frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}} - \frac{9}{8} \log | (x + \frac{3}{2}) + \sqrt{{(x + \frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}} | + C \\ = \frac{(2 x + 3)}{4} \sqrt{x^{2} + 3 x} - \frac{9}{8} \log | (x + \frac{3}{2}) + \sqrt{x^{2} + 3 x} | + C \end{array}$

Q.92

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

Ans.

$\begin{array}{l} We have, \sqrt{1 + \frac{x^{2}}{9}} = \frac{1}{3} \sqrt{9 + x^{2}} \\ Let \int \frac{1}{3} \sqrt{9 + x^{2}} dx = \frac{1}{3} \int \sqrt{3^{2} + x^{2}} dx \\ = \frac{1}{3} \int \sqrt{x^{2} + 3^{2}} dx \\ [∵ \int \sqrt{x^{2} + a^{2}} dx = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} |] \\ = \frac{1}{3} (\frac{x}{2} \sqrt{x^{2} + 3^{2}} + \frac{3^{2}}{2} \log | x + \sqrt{x^{2} + 3^{2}} |) + C \\ = \frac{x}{6} \sqrt{x^{2} + 9} + \frac{3}{2} \log | x + \sqrt{x^{2} + 9} | + C \end{array}$

Q.93

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

Ans.

$\begin{array}{l} Let I = \int \sqrt{1 + x^{2}} dx \\ = \int \sqrt{x^{2} + 1^{2}} dx \\ [∵ \int \sqrt{x^{2} + a^{2}} dx = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} |] \\ = (\frac{x}{2} \sqrt{x^{2} + 1^{2}} + \frac{1^{2}}{2} \log | x + \sqrt{x^{2} + 1^{2}} |) + C \\ = \frac{x}{2} \sqrt{x^{2} + 1} + \frac{1}{2} \log | x + \sqrt{x^{2} + 1} | + C \\ = \frac{x}{2} \sqrt{1 + x^{2}} + \frac{1}{2} \log | x + \sqrt{1 + x^{2}} | + C \\ Hence, the correct option is A. \end{array}$

Q.94

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

Ans.

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

Q.95

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

Ans.

$\begin{array}{l} \int x \sqrt{x + x^{2}} dx \\ Let x = A \frac{d}{dx} (x + x^{2}) + B \\ x = A (1 + 2 x) + B \\ \Rightarrow 1 = 2 A & A + B = 0 \\ \Rightarrow A = \frac{1}{2} & B = - \frac{1}{2} \\ ∴ \int x \sqrt{x + x^{2}} dx = \frac{1}{2} \int (1 + x) \sqrt{x + x^{2}} dx - \frac{1}{2} \int \sqrt{x + x^{2}} dx \\ = \frac{1}{2} \times \frac{{(x + x^{2})}^{\frac{3}{2}}}{(\frac{3}{2})} - \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} dx \\ = \frac{1}{3} \times {(x + x^{2})}^{\frac{3}{2}} - \frac{1}{2} \times \frac{(x + \frac{1}{2})}{2} \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} \\ + \frac{1}{2} \times \frac{{(\frac{1}{2})}^{2}}{2} \log | x + \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} | + C \\ = \frac{{(x + x^{2})}^{\frac{3}{2}}}{3} - \frac{1}{8} (2 x + 1) \sqrt{x + x^{2}} + \frac{1}{16} \log | x + \sqrt{x + x^{2}} | + C \end{array}$

Q.96

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

Ans.

$\begin{array}{l} \int (x + 1) \sqrt{2 x^{2} + 3} dx \\ Let x + 1 = A \frac{d}{dx} (2 x^{2} + 3) + B \\ x + 1 = A (4 x + 0) + B \\ \Rightarrow x + 1 = 4 Ax & B = 1 \\ \Rightarrow A = \frac{1}{4} & B = 1 \\ ∴ \int (x + 1) \sqrt{2 x^{2} + 3} dx = \frac{1}{4} \int 4 x \sqrt{2 x^{2} + 3} dx + 1 \int \sqrt{2 x^{2} + 3} dx \\ = \frac{1}{4} \times \frac{{(2 x^{2} + 3)}^{\frac{3}{2}}}{\frac{3}{2}} + \sqrt{2} \int \sqrt{x^{2} + {(\sqrt{\frac{3}{2}})}^{2}} dx \\ = \frac{1}{6} \times {(2 x^{2} + 3)}^{\frac{3}{2}} + \sqrt{2} \int \sqrt{x^{2} + {(\sqrt{\frac{3}{2}})}^{2}} dx \\ = \frac{1}{6} \times {(2 x^{2} + 3)}^{\frac{3}{2}} + \sqrt{2} (\begin{array}{l} \frac{1}{2} x \sqrt{x^{2} + {(\sqrt{\frac{3}{2}})}^{2}} \\ + \frac{{(\sqrt{\frac{3}{2}})}^{2}}{2} \log | x + {(\sqrt{\frac{3}{2}})}^{2} | \end{array}) + C \\ = \frac{1}{6} \times {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{\sqrt{2}}{2} x \sqrt{x^{2} + \frac{3}{2}} \\ + \frac{3 \sqrt{2}}{4} \log | x + \sqrt{x^{2} + {(\sqrt{\frac{3}{2}})}^{2}} | + C \\ = \frac{{(2 x^{2} + 3)}^{\frac{3}{2}}}{6} - \frac{x}{2} \sqrt{2 x^{2} + 3} - \frac{3 \sqrt{2}}{4} \log | x + \sqrt{x^{2} + \frac{3}{2}} | + C \end{array}$

Q.97

$(x + 3) \sqrt{3 - 4 x - x^{2}}$

Ans.

$\begin{array}{l} \int (x + 3) \sqrt{3 - 4 x - x^{2}} dx \\ Let x + 3 = A \frac{d}{dx} (3 - 4 x - x^{2}) + B \\ x + 3 = A (- 4 - 2 x) + B \\ \Rightarrow 3 = - 4 A + B & - 2 A = 1 \\ \Rightarrow A = \frac{1}{- 2} \\ & B = 3 + 4 A \\ = 3 + 4 (\frac{1}{- 2}) \\ = 3 - 2 = 1 \\ ∴ \int (x + 3) \sqrt{3 - 4 x - x^{2}} dx = \frac{1}{- 2} \int (- 4 - 2 x) \sqrt{3 - 4 x - x^{2}} dx \\ + 1 \int \sqrt{3 - 4 x - x^{2}} dx \\ = \frac{1}{- 2} \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{\frac{3}{2}} + \int \sqrt{7 - {(x + 2)}^{2}} dx \\ = \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{- 3} + \int \sqrt{{(\sqrt{7})}^{2} - {(x + 2)}^{2}} dx \\ = \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{- 3} + \frac{1}{2} (x + 2) \sqrt{{(\sqrt{7})}^{2} - {(x + 2)}^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C \\ + \frac{{(\sqrt{7})}^{2}}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C \\ = \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{- 3} + \frac{1}{2} (x + 2) \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C \end{array}$

Q.98

$Evaluate the following definite integrals as limit of sums . \int_{a}^{b} x dx$

Ans.

$\begin{array}{l} By definition \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, f (x) = x, h = \frac{b - a}{n} \\ ∴ \int_{a}^{b} x dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [a + a + h + . .. + a + (n - 1) h] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [na + h + 2 h + . .. + (n - 1) h] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [na + {1 + 2 + . .. + (n - 1)} h] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [na + \frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = (b - a) \lim_{n \to \infty} [a + \frac{(n - 1)}{2} (\frac{b - a}{n})] \\ = (b - a) \lim_{n \to \infty} [a + \frac{(b - a)}{2} (1 - \frac{1}{n})] \\ = (b - a) [a + \frac{(b - a)}{2} (1 - 0)] \\ = (b - a) (\frac{2 a + b - a}{2}) \\ = (b - a) \frac{(b + a)}{2} = \frac{b^{2} - a^{2}}{2} \end{array}$

Q.99

$Evaluate the following definite integrals as limit of sums . \int_{0}^{5} (x + 1) dx$

Ans.

$\begin{array}{l} We have, \int_{0}^{5} (x + 1) dx \\ By definition \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, a = 0, b = 5, f (x) = x + 1, h = \frac{5 - 0}{n} = \frac{5}{n} \\ ∴ \int_{0}^{5} (x + 1) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (0 + h) + . .. + f (0 + (n - 1) h)] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [(0 + 1) + (h + 1) + . .. + {(n - 1) h + 1}] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [n + h + 2 h + . .. + (n - 1) h] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [n + {1 + 2 + . .. + (n - 1)} h] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = 5 \lim_{n \to \infty} [1 + \frac{(n - 1)}{2} (\frac{5}{n})] \\ = 5 \lim_{n \to \infty} [1 + \frac{5}{2} (1 - \frac{1}{n})] \\ = 5 [1 + \frac{5}{2} (1 - 0)] \\ = 5 \times \frac{7}{2} = \frac{35}{2} \end{array}$

Q.100

$Evaluate the following definite integrals as limit of sums . \int_{2}^{3} x^{2} dx$

Ans.

$\begin{array}{l} We have, \int_{2}^{3} x^{2} dx \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, a = 2, b = 3, f (x) = x^{2}, h = \frac{3 - 2}{n} = \frac{1}{n} \\ ∴ \int_{2}^{3} x^{2} dx \\ = (3 - 2) \lim_{n \to \infty} \frac{1}{n} [f (2) + f (2 + h) + . .. + f (2 + (n - 1) h)] \\ = \lim_{n \to \infty} \frac{1}{n} [{(2)}^{2} + {(2 + h)}^{2} + . .. + {2 + (n - 1) h}^{2}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 + (4 + 4 h + h^{2}) + . .. + {4 + 4 (n - 1) h + {(n - 1)}^{2} h^{2}}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 n + 4 {1 + 2 + . .. + (n - 1)} h + {1 + 2^{2} + 3^{2} + . .. + {(n - 1)}^{2}} h^{2}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 n + 4 \frac{(n - 1) n}{2} h + \frac{(n - 1) n (2 n - 2 + 1)}{6} h^{2}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 n + 4 \frac{(n - 1) n}{2} . \frac{1}{n} + \frac{(n - 1) n (2 n - 2 + 1)}{6} . \frac{1}{n^{2}}] \\ [∵ \sum n = \frac{n (n + 1)}{2}, \sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] \\ = \lim_{n \to \infty} [4 + 4 \frac{(1 - \frac{1}{n})}{2} + \frac{(1 - \frac{1}{n}) (2 - \frac{1}{n})}{6}] \\ = [4 + 4 \frac{(1 - 0)}{2} + \frac{(1 - 0) (2 - 0)}{6}] \\ = 4 + 2 + \frac{1}{3} \\ = \frac{19}{3} \end{array}$

Q.101

$Evaluate the following definite integrals as limit of sums . \int_{1}^{4} (x^{2} - x) dx$

Ans.

$\begin{array}{l} We have, \int_{1}^{4} (x^{2} - x) dx = \int_{1}^{4} x^{2} dx - \int_{1}^{4} x dx \\ = I_{1} - I_{2} (Let) \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)] \\ where, h = \frac{b - a}{n} \\ Here, a = 1, b = 4, f (x) = x^{2}, h = \frac{4 - 1}{n} = \frac{3}{n} \\ ∴ I_{1} = \int_{1}^{4} x^{2} dx \\ = (4 - 1) \lim_{n \to \infty} \frac{1}{n} [f (1) + f (1 + h) + . .. + f (1 + (n - 1) h)] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [{(1)}^{2} + {(1 + h)}^{2} + . .. + {1 + (n - 1) h}^{2}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [1 + (1 + 2 h + h^{2}) + . .. + {1 + 2 (n - 1) h + {(n - 1)}^{2} h^{2}}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + {1 + 2 + . .. + (n - 1)} 2 h + {1 + 2^{2} + 3^{2} + . .. + {(n - 1)}^{2}} h^{2}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) n}{2} 2 h + \frac{(n - 1) n (2 n - 2 + 1)}{6} h^{2}] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + (n - 1) n . \frac{3}{n} + \frac{(n - 1) n (2 n - 2 + 1)}{6} . \frac{9}{n^{2}}] \\ [∵ \sum n = \frac{n (n + 1)}{2}, \sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] \\ = 3 \lim_{n \to \infty} [1 + (1 - \frac{1}{n}) \times 3 + \frac{(1 - \frac{1}{n}) (2 - \frac{1}{n})}{6} \times 9] \end{array}$

$\begin{array}{l} = 3 [1 + 3 (1 - 0) + \frac{3 (1 - 0) (2 - 0)}{2}] \\ = 3 (1 + 3 + 3) \\ = 21 \\ For { I}_{2} = \int_{1}^{4} x dx \\ ∴ \int_{1}^{4} x dx = (4 - 1) \lim_{n \to \infty} \frac{1}{n} [f (1) + f (1 + h) + . .. + f (1 + (n - 1) h)] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [1 + 1 + h + . .. + 1 + (n - 1) h] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + h + 2 h + . .. + (n - 1) h] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + {1 + 2 + . .. + (n - 1)} h] \\ = 3 \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = 3 \lim_{n \to \infty} [1 + \frac{(n - 1)}{2} (\frac{3}{n})] \\ = 3 \lim_{n \to \infty} [1 + \frac{3}{2} (1 - \frac{1}{n})] \\ = 3 [1 + \frac{3}{2} (1 - 0)] \\ = 3 (\frac{5}{2}) = \frac{15}{2} \\ So, \int_{1}^{4} (x^{2} - x) dx \\ = I_{1} - I_{2} = 21 - \frac{15}{2} = \frac{27}{2} \end{array}$

Q.102

$Evaluate the following definite integrals as limit of sums . \int_{- 1}^{1} e^{x} dx$

Ans.

$\begin{array}{l} We have, \int_{- 1}^{1} e^{x} dx \\ ∵ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, a = - 1, b = 1, f (x) = e^{x}, h = \frac{1 - (- 1)}{n} = \frac{2}{n} \\ ∴ \int_{- 1}^{1} e^{x} dx = (1 + 1) \lim_{n \to \infty} \frac{1}{n} [f (- 1) + f (- 1 + h) + . .. + f (- 1 + (n - 1) h)] \\ = 2 \lim_{n \to \infty} \frac{1}{n} [e^{- 1} + e^{- 1 + h} + . .. + e^{- 1 + (n - 1) h}] \\ Using the sum to n terms of a G.P., where a = e^{- 1}, r = e^{h}, \\ we have \\ ∴ \int_{- 1}^{1} e^{x} dx = 2 \lim_{n \to \infty} \frac{1}{n} [\frac{e^{- 1} (e^{nh} - 1)}{(e^{h} - 1)}] \\ = 2 \lim_{n \to \infty} \frac{1}{n} [\frac{e^{- 1} (e^{n . \frac{2}{n}} - 1)}{(e^{\frac{2}{n}} - 1)}] \\ = 2 \lim_{n \to \infty} \frac{1}{n} [\frac{e^{- 1} (e^{2} - 1)}{(e^{\frac{2}{n}} - 1)}] \\ = \frac{2 e^{- 1} (e^{2} - 1)}{\lim_{n \to \infty} [\frac{e^{\frac{2}{n}} - 1}{\frac{2}{n}}] . 2} \\ = \frac{2 e^{- 1} (e^{2} - 1)}{2} \\ = e - \frac{1}{e} [∵ \lim_{x \to 0} \frac{e^{h} - 1}{h} = 1] \end{array}$

Q.103

$Evaluate the following definite integrals as limit of sums . \int_{0}^{4} (x + e^{2 x}) dx$

Ans.

$\begin{array}{l} We have, \int_{0}^{4} (x + e^{2 x}) dx = \int_{0}^{4} x dx + \int_{0}^{4} e^{2 x} dx \\ = I_{1} + I_{2} (Let) \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)] \\ where, h = \frac{b - a}{n} \\ Here, a = 0, b = 4, f (x) = x, h = \frac{4 - 0}{n} = \frac{4}{n} \\ I_{1} = \int_{0}^{4} x dx \\ ∴ \int_{0}^{4} x dx = (4 - 0) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (0 + h) + . .. + f (0 + (n - 1) h)] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [0 + h + . .. + (n - 1) h] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [h + 2 h + . .. + (n - 1) h] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [{1 + 2 + . .. + (n - 1)} h] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = 4 \lim_{n \to \infty} [\frac{(n - 1)}{2} (\frac{4}{n})] \\ = 4 \lim_{n \to \infty} [\frac{4}{2} (1 - \frac{1}{n})] \\ = 4 [2 (1 - 0)] \\ = 4 (2) = 8 \\ I_{2} = \int_{0}^{4} e^{2 x} dx \\ = (4 - 0) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (0 + h) + . .. + f (0 + (n - 1) h)] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [e^{0} + e^{2 h} + . .. + e^{2 (n - 1) h}] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [1 + e^{2 h} + . .. + e^{2 (n - 1) h}] \\ Using the sum to n terms of a G.P., where a = 1, r = e^{2 h}, \\ we have \end{array}$

$\begin{array}{l} ∴ \int_{0}^{4} e^{2 x} dx = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{1 . (e^{2 nh} - 1)}{(e^{2 h} - 1)}] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{(e^{2 n . \frac{4}{n}} - 1)}{(e^{\frac{8}{n}} - 1)}] \\ = 4 \lim_{n \to \infty} \frac{1}{n} [\frac{(e^{8} - 1)}{(e^{\frac{8}{n}} - 1)}] \\ = \frac{4 (e^{8} - 1)}{\lim_{n \to \infty} [\frac{e^{\frac{8}{n}} - 1}{\frac{8}{n}}] . 8} \\ = \frac{4 (e^{8} - 1)}{8} [∵ \lim_{x \to 0} \frac{e^{h} - 1}{h} = 1] \\ = \frac{e^{8} - 1}{2} \\ So, \int_{0}^{4} (x + e^{2 x}) dx \\ = I_{1} + I_{2} \\ = 8 + \frac{e^{8} - 1}{2} \\ = \frac{15 + e^{8}}{2} \end{array}$

Q.104

$\int_{- 1}^{1} (x + 1) dx$

Ans.

$\begin{array}{l} Let I = \int_{- 1}^{1} (x + 1) dx \\ ∵ \int (x + 1) dx = \frac{x^{2}}{2} + x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (1) - F (- 1) \\ = (\frac{1^{2}}{2} + 1) - {\frac{{(- 1)}^{2}}{2} + (- 1)} \\ = \frac{3}{2} - (- \frac{1}{2}) \\ = 2 \end{array}$

Q.105

$\int_{2}^{3} \frac{1}{x} dx$

Ans.

$\begin{array}{l} Let I = \int_{2}^{3} \frac{1}{x} dx \\ ∵ \int \frac{1}{x} dx = logx = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (3) - F (2) \\ = \log 3 - \log 2 = \log \frac{3}{2} \end{array}$

Q.106

$\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) dx$

Ans.

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

Q.107

$\int_{0}^{\frac{π}{4}} \sin 2 x dx$

Ans.

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{4}} \sin 2 x dx \\ ∵ \int \sin 2 x dx = - \frac{1}{2} \cos 2 x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{4}) - F (0) \\ = - \frac{1}{2} \cos 2 (\frac{π}{4}) - (- \frac{1}{2} \cos 0) \\ = - \frac{1}{2} \cos \frac{π}{2} + \frac{1}{2} \cos 0 \\ = 0 + \frac{1}{2} (1) = \frac{1}{2} \end{array}$

Q.108

$\int_{0}^{\frac{π}{2}} \cos 2 x dx$

Ans.

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \cos 2 x dx \\ ∵ \int \cos 2 x dx = \frac{1}{2} \sin 2 x = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (\frac{π}{2}) - F (0) \\ = - \frac{1}{2} \sin 2 (\frac{π}{2}) - (- \frac{1}{2} \sin 0) \\ = - \frac{1}{2} sinπ + \frac{1}{2} \sin 0 \\ = 0 + 0 = 0 \end{array}$

Q.109

$\int_{4}^{5} e^{x} dx$

Ans.

$\begin{array}{l} Let I = \int_{4}^{5} e^{x} dx \\ ∵ \int e^{x} dx = e^{x} = F (x) \\ Therefore, by the second fundamental theorem, we have \\ I = F (5) - F (4) \\ = e^{5} - e^{4} \\ = e^{4} (e - 1) \end{array}$

Q.110

$\int_{0}^{\frac{π}{4}} \tan x dx$

Ans.

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{4}} \tan x dx \\ ∵ \int \tan x dx = - \log | cosx \end{array}$

NCERT Solutions Class 12 Mathematics Chapter 7

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