# NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.6) Exercise 7.6

The scientific discipline of mathematics has a long history that spans ages and the entire world. Mathematics has an immeasurable potential for academic and scientific research. This is also the reason that it is a field with a wide variety of career choices and opportunities that are quite lucrative and intriguing in themselves. Mathematics is a subject that is of vital importance in our daily lives. It is quintessential for the unhindered operation of a majority of human activities and human-made apparatuses. Mathematics is a prominent subject within the NCERT-prescribed curriculum for Class 12. This underlines the importance of access to reliable and easy-to-understand NCERT Solutions to aid and encourage the process of self-learning among students. The NCERT curriculum for Class 12 is made up of thirteen different chapters, which have been compiled into a comprehensively ordered sequence. The crucial theme of Integrals has been covered in the seventh chapter. Exercise 7.6 is a comprehensive and challenging assessment that is a prominent part of the chapter on Integrals. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 have been compiled with painstaking efforts in collaboration with knowledgeable subject experts by the Extramarks’ learning platform.

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## NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.6) Exercise 7.6

It has often been seen that students harbour an apprehensive attitude toward complex themes like Integrals and Calculus. However, these themes are an inseparable part of the NCERT academic curriculum prescribed for Class 12. The compilation of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 has been done meticulously by Extramarks to ensure that they are comprehensive and easy to comprehend. The cautiously-ordered framework of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 would be of great help to students in enhancing their conceptual clarity about the complex topics facing them. Integrals are not only a prominent part of the academic curriculum for Class 12, but they are also a required field of study for those who desire to pursue a career in mathematics. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 provided by the Extramarks’ website, along with adequate guidance from teachers, would be undoubtedly helpful to students in passing their examinations with flying colours.

### List of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 7- Integrals

A variety of topics associated with Integrals have been covered under the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6. The seventh chapter starts with an introduction to Integrals. Further, this chapter describes the two major types of Integrals, namely, the Definite and the Indefinite Integrals. Calculus can be defined as the theorem that makes up the connection between these two types of integrals. A multitude of other associated themes has also been covered in great detail in this chapter. These include Integration as an inverse process of differentiation, geometrical interpretation of indefinite integral, properties of indefinite integrals and comparison between differentiation and integration. The Extramarks’ website provides the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6, which cover all of these topics in minute detail. These notes are easy-to-follow and logically organized to ensure maximum conceptual clarity.

In the Theorem of Calculus, “integration by parts” refers to a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. Extramarks grants the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 which covers the topic of integration by parts in detail along with its practical application to solve questions.

### Access NCERT Solution for Class 12 Chapter 7 – Integrals

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### What is There in Exercise 7.6 Class 12 NCERT Solutions?

It can be said with a certain amount of certainty that within the entirety of the academic syllabus of NCERT prescribed for Mathematics for Class 12, Integrals is one of the most complex topics. The various underlying themes associated with Integrals have to be understood both at the level of individual formulae and as a comprehensive whole of a fundamental unit of Mathematics. It may often prove difficult for students to match the pace of instruction in the classroom or to grasp the content being taught at the same pace as their peers. It is also quite a cumbersome task to effectively accommodate all doubts and queries of students and to respond to them appropriately in the limited hours of classroom teaching. However, the conceptual clarity of students can take a toll if they are not able to receive the instruction being delivered by their teachers or have unresolved queries. Therefore, Extramarks is providing crucial learning resources like the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 to complement the guidance of teachers as well as the sincere efforts of students. The well-organized structure of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 compiled by Extramarks can be utilized by students to clarify their doubts by themselves through self-learning and regular practice.

### What is Integration by Parts of Ex 7.6?

Integration by parts, also referred to as “partial integration”, is the name of a process wherein the integral of a product of functions is found in terms of the integral of the product of their derivative and antiderivative. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 cover this topic in greater detail.

### How does Class 12 Maths Chapter 7 Exercise 7.6 Help Students?

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 have been compiled in a format that makes them easier to understand and retain. They also abide by the logical sequence of calculations. The format of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 is compiled by the Extramarks’ learning platform. This would also help the students acquire the comprehension skills necessary for figuring out the type and number of formulae they would have to apply to solve a particular question.

### Solved Examples

Extramarks is committed to providing reliable and authentic, high-quality online versions of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 to aid the sincere efforts of students. The solutions covered within the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6 have been prepared in careful consideration of the NCERT syllabus and the conventional CBSE examination pattern. This has been done to ensure that students have clarity regarding the kinds of answers they are expected to write in exams to secure perfect scores.

Q.1 Integrate the function x sin x.

Ans

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

Q.2 Integrate the functions
x sin 3x

Ans

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

Q.3 Integrate the functions
x2 ex

Ans

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

Q.4 Integrate the functions
x logx

Ans

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

Q.5 Integrate the functions
x log 2x

Ans

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

Q.6 Integrate the functions
x2 logx

Ans

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

Q.7 Integrate the functions
x sin-1x

Ans

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

Q.8 Integrate the functions
x tan-1x

Ans

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

Q.9 Integrate the functions
x cos-1x

Ans

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

Q.10 Integrate the functions
(sin–1x)2

Ans

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

Q.11

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

Ans

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

Q.12

Integrate the functions x sec2x

Ans

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

Q.13

Integrate the functions tan-1x

Ans

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

Q.14

Integrate the functions x (logx)2

Ans

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

Q.15

Integrate the functions (x2+1) logx

Ans

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

Q.16

Integrate the functions ex (sinx + cosx)

Ans

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

Q.17

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

Ans

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

Q.18

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

Ans

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

Q.19

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

Ans

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

Q.20

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

Ans

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

Q.21

Integrate the functions e2x sinx

Ans

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

Q.22

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

Ans

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

Q.23

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

Ans

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

Q.24

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

Ans

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