# NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT solutions for class 12 maths chapter 7 Integrals: The word integration literally means summation. When you have to find the sum of finite numbers you can do by simply adding these numbers. But when you are finding the sum of a certain number of elements as the number of elements tends to infinity and at the same time each term becomes infinitesimally small, you can use a prosses to find its limit called integration. In this article, you will find NCERT solutions for class 12 maths chapter 7 integrals which is very helpful when you are solving NCERT textbook questions. In the differential calculus you learnt about the differentiation, defining tangent and and how to calculate the slope of the line. Integration is the inverse process of differentiation. Remember every function is not integrable which means you can't integrate every function. The function is integrable only if the function is already differentiated. In the solutions of NCERT for class 12 maths chapter 7 integrals, you will learn the different methods of integration for different types of functions. CBSE NCERT solutions for class 12 maths chapter 7 integrals are very important from the 12th CBSE board exam point of view and are also important for competitive examinations like JEE Main, VITEEE, BITSAT, etc. Here you will find all NCERT solutions from class 6 to 12 at a single place to help you to learn CBSE maths and science.

Integrals has 13 % weightage in 12 board final examination. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own. If you are not able to do, you can take the help of these CBSE NCERT solutions for class 12 maths chapter 7 integrals. In this chapter, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercise. Here, the NCERT solutions for class 12 maths chapter 7 integrals are solved and explained in detail to develop a grip on the topic. Here, you will learn two types of integrals: Definite integral and  Indefinite integral and also learn their properties and formulas.

 Definite Integral Indefinite Integral Definition A definite Integral has upper and lower limits if  'a' and 'b' are the limits or boundaries. The definite integral of f(x) is a number, not function. An integral without upper limit and lower limit. It is also an antiderivative. The indefinite integral of f(x) is a function not number. Expression $\dpi{120} \int_{a}^{b}f(x)dx$ $F(x)=\int f(x)dx$

## 7.1 Introduction

7.2 Integration as an Inverse Process of Differentiation

7.2.1 Geometrical interpretation of indefinite integral

7.2.2 Some properties of indefinite integral

7.2.3 Comparison between differentiation and integration

7.3 Methods of Integration

7.3.1 Integration by substitution

7.3.2 Integration using trigonometric identities

7.4 Integrals of Some Particular Functions

7.5 Integration by Partial Fractions

7.6 Integration by Parts

7.7 Definite Integral

7.7.1 Definite integral as the limit of a sum

7.8 Fundamental Theorem of Calculus

7.8.1Area function

7.8.2 First fundamental theorem of integral calculus

7.8.3 Second fundamental theorem of integral calculus

7.9 Evaluation of Definite Integrals by Substitution

7.10 Some Properties of Definite Integrals

## NCERT Solutions for Class 12 Maths Chapter 7 Integrals- Exercise Questions

Solutions of NCERT for class 12 maths chapter 7 Integrals Exercise: 7.2

### Question:1 Integrate the functions  $\frac{2x}{1+ x ^2}$

Given to integrate $\frac{2x}{1+ x ^2}$ function,

Let us assume $1+x^2 =t$

we get, $2xdx = dt$

$\implies \int \frac{2x}{1+x^2} dx = \int \frac{1}{t} dt$

$= \log|t| +C$

$= \log|1+x^2| +C$    now back substituting the value of $t = 1+x^2$

as $(1+x^2)$ is positive we can write

$= \log(1+x^2) +C$

### Question:2 Integrate the functions $\frac{( \log x )^2}{x}$

Given to integrate $\frac{( \log x )^2}{x}$ function,

Let us assume $\log |x| = t$

we get, $\frac{1}{x}dx= dt$

$\implies \int \frac{\left ( \log|x| \right )^2}{x}\ dx = \int t^2dt$

$= \frac{t^3}{3}+C$

$= \frac{(\log|x|)^3}{3}+C$

### Question:3  Integrate the functions $\frac{1}{x+ x \log x }$

Given to integrate $\frac{1}{x+ x \log x }$ function,

Let us assume $1+\log x = t$

we get, $\frac{1}{x}dx= dt$

$\implies \int \frac{1}{x(1+\log x )} dx = \int \frac{1}{t} dt$

$= \log|t| +C$

$= \log |1+ \log x | +C$

### Question:4 Integrate the functions $\sin x \sin ( \cos x )$

Given to integrate $\sin x \sin ( \cos x )$ function,

Let us assume $\cos x =t$

we get, $-\sin x dx =dt$

$\implies \int \sin x \sin(\cos x)dx = -\int \sin t dt$

$= -\left ( -\cos t \right ) +C$

$= \cos t +C$

Back substituting the value of t we get,

$= \cos (\cos x ) +C$

### Question:5 Integrate the functions  $\sin ( ax + b ) \cos ( ax + b )$

Given to integrate $\sin ( ax + b ) \cos ( ax + b )$ function,

$\sin ( ax + b ) \cos ( ax + b ) = \frac{2\sin ( ax + b ) \cos ( ax + b )}{2} = \frac{\sin 2(ax+b)}{2}$

Let us assume $2(ax+b) = t$

we get, $2adx =dt$

$\int \frac{\sin 2(ax+b)}{2} dx = \frac{1}{2}\int \frac{\sin t}{2a} dt$

$= \frac{1}{4a}[-cos t] +C$

Now, by back substituting the value of t,

$= \frac{-1}{4a}[cos 2(ax+b)] +C$

### Question:6 Integrate the functions $\sqrt { ax + b }$

Given to integrate $\sqrt { ax + b }$ function,

Let us assume $(ax+b) = t$

we get, $adx =dt$

$dx = \frac{1}{a}dt$

$\Rightarrow \int(ax+b)^{\frac{1}{2}} dx = \frac{1}{a}\int t^{\frac{1}{2}}dt$

Now, by back substituting the value of t,

$= \frac{1}{a}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

$= \frac{2(ax+b)^\frac{3}{2}}{3a} +C$

### Question:7 Integrate the functions  $x \sqrt { x +2 }$

Given function $x \sqrt { x +2 }$,

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

Assume the $(x+2) = t$19634

$\therefore dx =dt$

$\Rightarrow \int x\sqrt{x+2} dx = \int (t-2) \sqrt{t} dt$

$= \int (t-2) \sqrt{t} dt$

$= \int \left ( t^{\frac{3}{2}}-2t^{\frac{1}{2}} \right )dt$

$= \int t^{\frac{3}{2}}dt -2\int t^{\frac{1}{2}}dt$

$= \frac{t^{\frac{5}{2}}}{\frac{5}{2}} -2\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

$= \frac{2}{5}t^{\frac{5}{2}} -\frac{4}{3}t^{\frac{3}{2}} +C$

Back substituting the value of t in the above equation.

or, $\frac{2}{5}(x+2)^{\frac{5}{2}}- \frac{4}{3}(x+2)^\frac{3}{2} +C$ , where C is any constant value.

### Question:8 Integrate the functions $x \sqrt { 1+ 2 x^2 }$

Given function  $x \sqrt { 1+ 2 x^2 }$,

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

Assume the $1+2x^2= t$

$\therefore 4xdx =dt$

$\Rightarrow \int x\sqrt{1+2x^2}dx = \int \frac{\sqrt {t}}{4} dt$

Or $= \frac{1}{4}\int t^{\frac{1}{2}} dt = \frac{1}{4}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

$= \frac{1}{6}(1+2x^2)^{\frac{3}{2}} +C$ , where C is any constant value.

### Question:9 Integrate the functions $( 4x +2 ) \sqrt { x ^ 2 + x + 1 }$

Given function  $( 4x +2 ) \sqrt { x ^ 2 + x + 1 }$,

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

Assume the $1+x+x^2 = t$

$\therefore (2x+1)dx =dt$

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

$= \int 2\sqrt {t}dt = 2\int \sqrt{t}dt$

$= 2\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C$

Now, back substituting the value of t in the above equation,

$= \frac{4}{3}(1+x+x^2)^{\frac{3}{2}} +C$, where C is any constant value.

### Question:10 Integrate the functions  $\frac{1}{x - \sqrt x }$

Given function  $\frac{1}{x - \sqrt x }$,

$\int \frac{1}{x - \sqrt x } dx$

Can be written in the form:

$\frac{1}{x - \sqrt x } = \frac{1}{\sqrt {x}(\sqrt{x}-1)}$

Assume the $(\sqrt{x}-1) =t$

$\therefore \frac{1}{2\sqrt{x}}dx =dt$

$\Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}dx = \int \frac{2}{t}dt$

$= 2\log|t| +C$

$= 2\log|\sqrt{x}-1| +C$, where C is any constant value.

### Question:11 Integrate the functions  $\frac{x }{ \sqrt{ x +4} }$, x > 0

Given function  $\frac{x }{ \sqrt{ x +4} }$,

$\int \frac{x }{ \sqrt{ x +4} }dx$

Assume the $x+4 =t$  so, $x =t-4$

$\therefore dx=dt$

$\int \frac{x}{\sqrt{x+4}}dx = \int \frac{t-4}{\sqrt{t}}dt$

$\int t^\frac{1}{2}dt -4\int t^{\frac{-1}{2}}dt$

$= \frac{2}{3}t^{\frac{3}{2}} - 4\left ( 2t^{\frac{1}{2}} \right )+C$

$= \frac{2}{3}(x+4)^{\frac{3}{2}} -16(x+4)^{\frac{1}{2}}+C$

, where C is any constant value.

### Question:12 Integrate the functions $( x ^3 - 1 ) ^{1/3} x ^ 5$

Given function  $( x ^3 - 1 ) ^{1/3} x ^ 5$,

$\int ( x ^3 - 1 ) ^{1/3} x ^ 5 dx$

Assume the $x^3-1 = t$

$\therefore 3x^2dx=dt$

$\implies \int(x^3-1)^{\frac{1}{3}} x^5 dx = \int (x^3-1)^{\frac{1}{3}}x^3.x^2dx$

$= \int t^{\frac{1}{3}}(t+1)\frac{dt}{3}$

$= \frac{1}{3} \int \left ( t^\frac{4}{3}+t^\frac{1}{3} \right )dt$

$= \frac{1}{3}\left [ \frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right ]+C$

$= \frac{1}{3}\left [ \frac{3}{7}t^{\frac{7}{3}}+\frac{3}{4}t^{\frac{4}{3}} \right ]+C$

$= \frac{1}{7}(x^3-1)^{\frac{7}{3}} + \frac{1}{4}(x^3-1)^{\frac{4}{3}} +C$ , where C is any constant value.

### Question:13  Integrate the functions  $\frac{x ^2 }{(2+3x^3)^3}$

Given function  $\frac{x ^2 }{(2+3x^3)^3}$,

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

Assume the $2+3x^3 =t$

$\therefore 9x^2dx=dt$

$\implies \int\frac{x^2}{(2+3x^2)}dx = \frac{1}{9}\int \frac{dt}{t^3}$

$= \frac{1}{9}\left ( \frac{t^{-2}}{-2} \right ) +C$

$= \frac{-1}{18}\left ( \frac{1}{t^2} \right )+C$

$= \frac{-1}{18(2+3x^3)^2}+C$  , where C is any constant value.

### Question:14  Integrate the functions  $\frac{1}{x (\log x )^m} , x > 0 , m \neq 1$

Given function  $\frac{1}{x (\log x )^m} , x > 0 , m \neq 1$,

Assume the $\log x =t$

$\therefore \frac{1}{x}dx =dt$

$\implies \int\frac{1}{x(logx)^m}dx = \int\frac{dt}{t^m}$

$=\left ( \frac{t^{-m+1}}{1-m} \right ) +C$

$= \frac{(log x )^{1-m}}{(1-m)} +C$, where C is any constant value.

### Question:15  Integrate the functions  $\frac{x}{9- 4 x ^2 }$

Given function  $\frac{x}{9- 4 x ^2 }$,

Assume the $9-4x^2 =t$

$\therefore -8xdx =dt$

$\implies \int\frac{x}{9-4x^2} = -\frac{1}{8}\int \frac{1}{t}dt$

$= \frac{-1}{8}\log|t| +C$

Now back substituting the value of t ;

$= \frac{-1}{8}\log|9-4x^2| +C$ , where C is any constant value.

### Question:16  Integrate the functions $e ^{ 2 x +3 }$

Given function  $e ^{ 2 x +3 }$,

Assume the $2x+3 =t$

$\therefore 2dx =dt$

$\implies \int e^{2x+3} dx = \frac{1}{2}\int e^t dt$

$= \frac{1}{2}e^t +C$

Now back substituting the value of t ;

$= \frac{1}{2}e^{2x+3}+C$ , where C is any constant value.

### Question:17  Integrate the functions $\frac{x }{e^{x^{2}}}$

Given function  $\frac{x }{e^{x^{2}}}$,

Assume the $x^2=t$

$\therefore 2xdx =dt$

$\implies \int \frac{x}{e^{x^2}}dx = \frac{1}{2}\int \frac{1}{e^t}dt$

$= \frac{1}{2}\int e^{-t} dt$

$= \frac{1}{2}\left ( \frac{e^{-t}}{-1} \right ) +C$

$= \frac{-1}{2}e^{-x^2} +C$

$= \frac{-1}{2e^{x^2} }+C$, where C is any constant value.

### Question:18  Integrate the functions $\frac{e ^{\tan ^{-1}x}}{1+ x^2 }$

Given,

$\frac{e ^{\tan ^{-1}x}}{1+ x^2 }$

Let's do the following substitution

$\\ tan^{-1}x = t \\ \implies \frac{1}{1+x^2}dx = dt$

$\therefore \int \frac{e ^{\tan ^{-1}x}}{1+ x^2 }dx = \int e ^{t}dt = e^t + C$

$= e^{tan^{-1}x} + C$

### Question:19  Integrate the functions $\frac{e ^{2x}-1}{e ^{2x}+1}$

Given function  $\frac{e ^{2x}-1}{e ^{2x}+1}$,

Simplifying it by dividing both numerator and denominator by $e^x$, we obtain

$\frac{\frac{e^{2x}-1}{e^x}}{\frac{e^{2x}+1}{e^x}} = \frac{e^x-e^{-x}}{e^x+e^{-x}}$

Assume the $e^{x}+e^{-x} =t$

$\therefore (e^x-e^{-x})dx =dt$

$\implies \int \frac{e^{2x}-1}{e^{2x}+1}dx = \int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$

$= \int \frac{dt}{t}$

$= \log |t| +C$

Now, back substituting the value of t,

$= \log |e^x+e^{-x}| +C$ , where C is any constant value.

### Question:20 Integrate the functions $\frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }}$

Given function  $\frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }}$,

Assume the $e^{2x}+e^{-2x} =t$

$\therefore (2e^{2x}-2e^{-2x})dx =dt$

$\implies \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx = \int \frac{dt}{2t}$

$= \frac{1}{2}\int \frac{1}{t}dt$

$= \frac{1}{2}\log|t| +C$

Now, back substituting the value of t,

$= \frac{1}{2}\log|e^{2x}+e^{-2x}| +C$ , where C is any constant value.

### Question:21  Integrate the functions  $\tan ^2 ( 2x-3 )$

Given function  $\tan ^2 ( 2x-3 )$,

Assume the $2x-3 =t$

$\therefore 2dx =dt$

$\implies \int \tan^2(2x-3) dx = \frac{\int \tan^2(t)}{2}dt$

$=\frac{1}{2}\int (\sec^2t -1) dt$                      $\left [\because \tan^2t+1 = \sec^2 t \right ]$

$= \frac{1}{2}\left [ \tan t - t \right ] +C$

Now, back substituting the value of t,

$= \frac{1}{2}\left [ \tan(2x-3)-2x+3 \right ]+C$

or $\frac{1}{2} \tan(2x-3)-x+C$ , where C is any constant value.

### Question:22 Integrate the functions  $\sec ^2 ( 7- 4x )$

Given function  $\sec ^2 ( 7- 4x )$,

Assume the $7-4x=t$

$\therefore -4dx =dt$

$\implies \int \sec^2(7-4x)dx = \frac{-1}{4}\int \sec^2t dt$

$=-\frac{1}{4}(\tan t) +C$

Now, back substituted the value of t.

$=-\frac{1}{4}\tan(7-4x)+C$, where C is any constant value.

### Question:23  Integrate the functions $\frac{\sin ^{-1}x}{\sqrt { 1- x^2 }}$

Given function  $\frac{\sin ^{-1}x}{\sqrt { 1- x^2 }}$,

Assume the $\sin^{-1}x =t$

$\therefore \frac{1}{\sqrt{1-x^2}}dx = dt$

$\implies \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx =\int t dt$

$= \frac{t^2}{2}+C$

Now, back substituted the value of t.

$= \frac{(\sin^{-1}x)^2}{2}+C$, where C is any constant value.

### Question:24 Integrate the functions  $\frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x }$

Given function  $\frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x }$,

or simplified as $\frac{2 \cos x - 3\sin x }{2(3 \cos x + 2 \sin x) }$

Assume the $3\cos x +2\sin x =t$

$\therefore (-3\sin x + 2\cos x )dx =dt$

$\implies \int \frac{2\cos x - 3\sin x }{6\cos x +4\sin x }dx = \int \frac{dt}{2t}$

$= \frac{1}{2}\int \frac{dt}{t}$

$= \frac{1}{2}\log|t| +C$

Now, back substituted the value of t.

$= \frac{1}{2}\log|3\cos x +2\sin x| +C$, where C is any constant value.

### Question:25  Integrate the functions $\frac{1 }{ \cos ^2 x (1-\tan x )^2}$

Given function  $\frac{1 }{ \cos ^2 x (1-\tan x )^2}$,

or simplified as $\frac{1 }{ \cos ^2 x (1-\tan x )^2} = \frac{\sec^2x}{(1-\tan x)^2}$

Assume the $(1-\tan x)=t$

$\therefore -\sec^2xdx =dt$

$\implies \int \frac{\sec^2x}{(1-\tan x)^2}dx = \int\frac{-dt}{t^2}$

$= -\int t^{-2} dt$

$= \frac{1}{t} +C$

Now, back substituted the value of t.

$= \frac{1}{1-\tan x}+C$

where C is any constant value.

### Question:26  Integrate the functions $\frac{\cos \sqrt x }{\sqrt x }$

Given function  $\frac{\cos \sqrt x }{\sqrt x }$,

Assume the $\sqrt x =t$

$\therefore \frac{1}{2\sqrt x}dx =dt$

$\implies \int \frac{\cos \sqrt{x}}{\sqrt{x}}dx = 2\int \cos t dt$

$= 2\sin t +C$

Now, back substituted the value of t.

$= 2\sin \sqrt{x}+C$, where C is any constant value.

### Question:27  Integrate the functions $\sqrt { \sin 2x } \cos 2x$

Given function  $\sqrt { \sin 2x } \cos 2x$,

Assume the $\sin 2x = t$

$\therefore 2\cos 2x dx =dt$

$\implies \int \sqrt{\sin 2x }\cos 2x dx = \frac{1}{2}\int \sqrt t dt$

$= \frac{1}{2}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right )+C$

$= \frac{1}{3}t^{\frac{3}{2}}+C$

Now, back substituted the value of t.

$= \frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C$ , where C is any constant value.

### Question:28 Integrate the functions $\frac{\cos x }{\sqrt { 1+ \sin x }}$

Given function  $\frac{\cos x }{\sqrt { 1+ \sin x }}$,

Assume the $1+\sin x =t$

$\therefore \cos x dx = dt$

$\implies \int \frac{\cos x }{\sqrt{1+\sin x}}dx = \int \frac{dt}{\sqrt t}$

$= \frac{t^{\frac{1}{2}}}{\frac{1}{2}} +C$

$= 2\sqrt t +C$

Now, back substituted the value of t.

$= 2{\sqrt{1+\sin x}} +C$ , where C is any constant value.

### Question:29  Integrate the functions $\cot x \: log \sin x$

Given function  $\cot x \: log \sin x$,

Assume the $\log \sin x =t$

$\therefore \frac{1}{\sin x }.\cos x dx =dt$

$\cot x dx =dt$

$\implies \int \cot x \log \sin x dx =\int t dt$

$= \frac{t^2}{2}+C$

Now, back substituted the value of t.

$= \frac{1}{2}(\log \sin x )^2+C$, where C is any constant value.

### Question:30 Integrate the functions  $\frac{\sin x }{1+ \cos x }$

Given function  $\frac{\sin x }{1+ \cos x }$,

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

$\implies \int \frac{\sin x}{1+\cos x}dx = \int -\frac{dt}{t}$

$= -\log|t| +C$

Now, back substituted the value of t.

$= -\log|1+\cos x | +C$, where C is any constant value.

### Question:31  Integrate the functions $\frac{\sin x }{( 1+ \cos x )^2}$

Given function  $\frac{\sin x }{( 1+ \cos x )^2}$,

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

$\implies \int \frac{\sin x}{(1+\cos x)^2}dx = \int -\frac{dt}{t^2}$

$= -\int t^{-2}dt$

$= \frac{1}{t}+C$

Now, back substituted the value of t.

$= \frac{1}{1+\cos x} +C$, where C is any constant value.

### Question:32 Integrate the functions  $\frac{1}{1+ \cot x }$

Given function  $\frac{1}{1+ \cot x }$

Assume that $I = \int \frac{1}{1+ \cot x } dx$

Now solving the assumed integral;

$I = \int \frac{1}{1+ \frac{\cos x }{\sin x} } dx$

$= \int \frac{\sin x }{\sin x + \cos x } dx$

$= \frac{1}{2}\int \frac{2\sin x }{\sin x + \cos x } dx$

$= \frac{1}{2}\int \frac{(\sin x+ \cos x ) +(\sin x -\cos x ) }{(\sin x + \cos x) } dx$

$=\frac{1}{2}\int 1 dx + \frac{1}{2} \int \frac{\sin x -\cos x }{\sin x +\cos x } dx$

$=\frac{1}{2}(x) + \frac{1}{2} \int \frac{\sin x -\cos x }{\sin x +\cos x } dx$

Now, to solve further we will assume $\sin x + \cos x =t$

Or, $(\cos x -\sin x)dx =dt$

$\therefore I = \frac{x}{2}+ \frac{1}{2}\int \frac{-(dt)}{t}$

$= \frac{x}{2}- \frac{1}{2}\log|t| +C$

Now, back substituting the value of t,

$= \frac{x}{2}- \frac{1}{2}\log|\sin x + \cos x| +C$

### Question:33 Integrate the functions  $\frac{1}{1- \tan x }$

Given function  $\frac{1}{1- \tan x }$

Assume that $I = \int \frac{1}{1- \tan x } dx$

Now solving the assumed integral;

$I = \int \frac{1}{1-\frac{\sin x}{\cos x }} dx$

$= \int \frac{\cos x }{\cos x - \sin x } dx$

$= \frac{1}{2}\int \frac{2\cos x }{\cos x - \sin x } dx$

$= \frac{1}{2}\int \frac{(\cos x -\sin x ) +(\cos x +\sin x ) }{(\cos x - \sin x) } dx$

$=\frac{1}{2}\int 1 dx + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx$

$=\frac{1}{2}(x) + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx$

Now, to solve further we will assume $\cos x - \sin x =t$

Or, $(-\sin x-\cos x )dx =dt$

$\therefore I = \frac{x}{2}+ \frac{1}{2}\int \frac{-(dt)}{t}$

$= \frac{x}{2}- \frac{1}{2}\log|t| +C$

Now, back substituting the value of t,

$= \frac{x}{2}- \frac{1}{2}\log|\cos x - \sin x| +C$

### Question:34  Integrate the functions  $\frac{\sqrt { \tan x } }{\sin x \cos x }$

Given function  $\frac{\sqrt { \tan x } }{\sin x \cos x }$

Assume that $I = \int \frac{\sqrt { \tan x } }{\sin x \cos x }dx$

Now solving the assumed integral;

Multiplying numerator and denominator by $\cos x$;

$I = \int \frac{\sqrt{\tan x }\times\cos x}{\sin x \cos x\times \cos x}dx$

$= \int \frac{\sqrt{\tan x }}{\tan x \cos^2 x } dx$

$= \int \frac{\sec^2 x }{\sqrt{\tan x }}dx$

Now, to solve further we will assume $\tan x =t$

Or, $\sec^2{x}dx =dt$

$\therefore I =\int \frac{dt}{\sqrt t}$

$=2\sqrt t +C$

Now, back substituting the value of t,

$= 2\sqrt{\tan x } +C$

### Question:35  Integrate the functions  $\frac{( 1+ \log x )^2}{x}$

Given function  $\frac{( 1+ \log x )^2}{x}$

Assume that $1+\log x =t$

$\therefore \frac{1}{x}dx =dt$

$= \int \frac{(1+\log x )^2}{x}dx = \int t^2 dt$

$= \frac{t^3}{3}+C$

Now, back substituting the value of t,

$= \frac{(1+\log x )^3}{3}+C$

### Question:36  Integrate the functions  $\frac{( x+1)( x+ \log x )^2}{x }$

Given function  $\frac{( x+1)( x+ \log x )^2}{x }$

Simplifying to solve easier;

$\frac{( x+1)( x+ \log x )^2}{x } = \left ( \frac{x+1}{x} \right )\left ( x+\log x \right )^2$

$=\left ( 1+\frac{1}{x} \right )\left ( x+\log x \right )^2$

Assume that $x+\log x =t$

$\therefore \left ( 1+\frac{1}{x} \right )dx = dt$

$= \int \left ( 1+\frac{1}{x} \right )\left ( x+\log x \right )^2 dx = \int t^2 dt$

$= \frac{t^3}{3}+C$

Now, back substituting the value of t,

$= \frac{(x+\log x )^3}{3}+C$

### Question:37  Integrate the functions  $\frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }$

Given function  $\frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }$

Assume that $x^4 =t$

$\therefore 4x^3 dx =dt$

$\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx = \frac{1}{4} \int \frac{\sin(\tan^{-1} t)}{1+t^2}dt$            ......................(1)

Now to solve further we take $\tan ^{-1} t = u$

$\therefore \frac{1}{1+t^2} dt =du$

So, from the equation (1), we will get

$\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx =\frac{1}{4}\int \sin u\ du$

$= \frac{1}{4}(-\cos u) +C$

Now back substitute the value of u,

$= \frac{-1}{4}\cos (\tan^{-1} t) +C$

and then back substituting the value of t,

$= \frac{-1}{4}\cos (\tan^{-1} x^4) +C$

### Question:38  Choose the correct answer $\int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx\: \: \: equals$

$(A) 10^x - x^{10} + C \\\\(B) 10^x + x^{10} + C\\\\ (C) (10^x - x^{10})^{-1} + C \\\\ (D) log (10^x + x^{10}) + C$

Given integral $\int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx$

Taking the denominator $x^{10} +10^x = t$

Now differentiating both sides we get

$\therefore \left ( 10x^9+10^x\log_{e}10 \right )dx = dt$

$\implies \int \frac{10x^9+10^x\log_{e}10}{x^{10}+10^x} dx = \int \frac{dt}{t}$

$= \log t +C$

Back substituting the value of t,

$= \log (x^{10}+10^x) +C$

Therefore the correct answer is D.

### Question:39  Choose the correct answer  $\int \frac{dx }{\sin ^ 2 x \cos ^2 x }\: \: \: equals$

$(A) \tan x + \cot x + C \\\\ (B) \tan x - \cot x + C\\\\ (C) \tan x \cot x + C\\\\ (D) \tan x - \cot 2x + C$

Given integral $\int \frac{dx }{\sin ^ 2 x \cos ^ 2x }$

$\int \frac{dx }{\sin ^ 2 x \cos ^ 2x } = \int \frac{1}{\sin ^2 x \cos ^2 x } dx$

$=\int \frac{\sin ^2 x +\cos^2 x }{\sin^2 x \cos^2 x}dx$                                         $\left ( \because \sin ^2 x +\cos^2 x =1 \right )$

$=\int \frac{\sin^2 x }{\sin^2 x \cos^2 x}dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x}dx$

$=\int \sec^2 x dx + \int cosec^2 x dx$

$=\tan x -\cot x +C$

Therefore, the correct answer is B.

CBSE NCERT solutions for class 12 maths chapter 7 Integrals Exercise: 7.3

### Question:1  Find the integrals of the functions $\sin ^ 2 ( 2x+ 5 )$

$\sin ^ 2 ( 2x+ 5 )$

using the trigonometric identity

$sin^2x=\frac{1-cos2x}{2}$

we can write the given question as

$\frac{1-\cos 2(2x+5)}{2}$$= \frac{1-\cos (4x+10)}{2}$
$\\=\int \frac{1-\cos (4x+10)}{2}dx\\ =\frac{1}{2}\int dx - \frac{1}{2}\int \cos(4x+10)dx\\ =\frac{x}{2}-\frac{1}{2}[\sin(4x+10)/4]\\ =\frac{x}{2}-\frac{\sin(4x+10)}{8}+C$

### Question:2  Find the integrals of the functions $\sin 3x \cos 4x$

Using identity $\sin A\cos B = 1/2[sin(A+B)+sin(A-B)]$

, therefore the given integral can be written as

$\int \sin 3x\cos 4x=\frac{1}{2}\int sin(3x+4x)+sin(3x-4x)\ dx$

$=\frac{1}{2}\int sin(7x)-sin(x)\ dx\\ =1/2[\int \sin (7x) dx-\int \sin x\ dx]\\ =\frac{1}{2}[(-1/7)\cos 7x+\cos x+ C]\\ = \frac{\cos x}{2}-\frac{\cos 7x}{14}+C$

### Question:3  Find the integrals of the functions $\cos 2x \cos 4x \cos 6x$

Using identity
$cosAcosB = \frac{1}{2}[cos(A+B)+cos(A-B)]$

$\int \cos 2x.\cos 4x.\cos 6x = \int \cos 2x. \frac{1}{2}[(\cos 10x)+\cos 2x]dx$

Again use the same identity mentioned in the first line

$\\= \frac{1}{2}\int (\cos 2x.\cos 10x+\cos 2x. \cos 2x)dx\\ =\frac{1}{2}\int\frac{1}{2}({\cos12x +\cos 8x})dx+\frac{1}{2}\int (\frac{1+\cos 4x}{2})dx\\ =\frac{\sin 12x}{48}+\frac{\sin 8x}{32}+\frac{\sin 4x}{16}+ x/4+C$

### Question:4 Find the integrals of the functions $\sin ^ 3 ( 2x +1 )$

$\int \sin^3(2x+1)dx = \int \sin^2(2x+1).\sin(2x+1)dx$

The integral can be written as

$= \int (1-\cos^2(2x+1)).\sin(2x+1)dx$
Let
$\\\cos (2x+1) =t\\ \sin (2x+1)dx = -dt/2$

$\\=\frac{-1}{2}\int (1-t^2)dt\\ =\frac{-1}{2}[t-t^3/3]\\ =\frac{t^3}{6}-\frac{t}{2}$

Now,  replace the value of t, we get;

$=\frac{\cos^3(2x+1)}{6}-\frac{\cos(2x+1)}{2}+C$

### Question:5 Find the integrals of the functions $\sin ^3 x \cos ^ 3 x$

$I = \int \sin^3x.\cos^3x\ dx$

rewrite the integral as follows

$\\=\int cos^3x.sin^2x.\sin x\ dx\\ =\int cos^3x(1-\cos^2x)\sin x\ dx$
Let   $\cos x = t \Rightarrow \sin x dx =-dt$

$\\=-\int t^3(1-t^2)dt\\ =-\int(t^3-t^5)dt\\ =-[\frac{t^4}{4}]+[\frac{t^6}{6}] +C\\ =\frac{\cos^6x}{6}-\frac{cos^4x}{4}+C$......(replace the value of t as $cos\ x$)

### Question:6 Find the integrals of the functions $\sin x \sin 2x \sin 3x$

Using the formula
$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the integral as follows

$\int \sin x.\sin 2x\sin 3x\ dx = \int \sin x\frac{1}{2}[\cos x-\cos 5x]dx$
$\\=\frac{1}{2} \int [\sin x.\cos x-\sin x.\cos 5x]dx\\ =\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x. \cos 5x\ dx\\ =-\frac{\cos 2x}{8}-\frac{1}{4}\int[\sin 6x -\sin 4x]\\ =-\frac{\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{6}+\frac{\cos 4x}{4}]\\ =-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}-\frac{\cos 4x}{16}+C$

### Question:7  Find the integrals of the functions $\sin 4x \sin 8x$

Using identity

$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the following integral as

$\sin 4x \sin 8x$ =
$\\=\frac{1}{2}\int(\cos 4x - \cos 12x) dx\\ =\frac{1}{2} [\int\cos 4x\ dx - \int \cos 12x\ dx]\\ =\frac{\sin 4x}{8}-\frac{\sin 12x}{24}+C$

### Question:8  Find the integrals of the functions $\frac{1- \cos x }{1+ \cos x }$

We know the identities

$\\1+\cos 2A = 2\cos^2A\\ 1-\cos 2A = 2\sin^2 A$

Using the above relations we can write

$\frac{1-\cos x}{1+\cos x}=\frac{\sin^2x/2}{\cos^2x/2} = \tan^2x/2$

$=\int \tan^2x/2 =\int (\sec^2x/2-1)dx$
$\\=\int (\sec^2x/2)dx-\int dx\\ = 2[\tan x/2]-{x}+C$

### Question:9 Find the integrals of the functions $\frac {\cos x }{1 + \cos x }$

The integral is rewritten using trigonometric identities

$\frac{\cos x}{1+ \cos x}= \frac{\cos^2x/2-\sin^2x/2}{2\cos^2x/2} =\frac{1}{2}[1-\tan^2x/2]$
$\\=\int \frac{1}{2}[1-\tan^2x/2] dx\\ =\frac{1}{2}\int 1-[sec^2\frac{x}{2}-1]=\frac{1}{2}\int 2-sec^2\frac{x}{2}\\=x-tan\frac{x}{2}+c$

### Question:10  Find the integrals of the functions $\sin ^ 4 x$

$\sin ^ 4 x$  can be written as follows using trigonometric identities
$\\=\sin^2x.\sin^2x\\ =\frac{1}{4}(1-\cos 2x)^{2}\\ =\frac{1}{4}(1+\cos^22x-2\cos 2x)\\ =\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)\\ =3/8+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}$

Therefore,
$\Rightarrow \int \sin^4x\ dx = \int \frac{3}{8}dx+\frac{1}{8}\int\cos 4x\ dx -\frac{1}{2}\int\cos 2x\ dx$
$= \frac{3x}{8}+\frac{\sin 4x}{32} -\frac{\sin 2x}{4}+C$

### Question:11  Find the integrals of the functions $\cos ^ 4 2x$

$cos^42x=cos^32xcos2x$

now using the identity

$cos^3x=\frac{cos3x+3cosx}{4}$

$cos^32xcos2x=\frac{cos6x +3cos2x}{4}cos2x=\frac{cos6xcos2x+3cos^22x}{4}$

now using the below two identities

$cosacosb=\frac{cos(a+b)+cos(a-b)}{2}\\and\ cos^22x=\frac{1+cos4x}{2}\\$

the value

$cos^42x=cos^32xcos2x\\=\frac{cos6xcos2x+3cos^22x}{4}=\frac{cos 4x+cos8x}{8}+\frac{3}{4}\frac{1+cos4x}{2}$.

the integral of the given function can be written as

$\int cos^42x=\int \frac{cos 4x+cos8x}{8}+\int \frac{3}{4}\frac{1+cos4x}{2}\\ \\=\frac{3}{8}x+\frac{sin4x}{8}+\frac{sin8x}{64}+C$

### Question:12 Find the integrals of the functions $\frac{\sin ^ 2x }{1+ \cos x }$

Using trigonometric identities we can write the given integral as follows.

$\frac{\sin ^ 2x }{1+ \cos x }$

$\\=\frac{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\\ =2\sin^2\frac{x}{2}\\ =1-\cos x$

$\therefore \int \frac{sin^22x}{1+\cos x} = \int (1-\cos x)dx$
$\\= \int 1dx-\int\cos x\ dx\\ =x-\sin x+C$

### Question:13  Find the integrals of the functions $\frac{\cos 2x - \cos 2 \alpha }{\cos x - \cos \alpha }$

We know that,

$\cos A-\cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$

Using this identity we can rewrite the given integral as

$\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\frac{-2\sin\frac{2x+2\alpha}{2}\sin\frac{2x-2\alpha}{2}}{-2\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}$

$\\=\frac{\sin(x+\alpha)\sin(x-\alpha)}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =\frac{[2\sin\frac{x+\alpha}{2}\cos \frac{x+\alpha}{2}][2\sin\frac{x-\alpha}{2}\cos\frac{x-\alpha}{2}]}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =4\cos\frac{x+\alpha}{2}\cos\frac{x-\alpha}{2}\\ =2[\cos x+\cos \alpha]$

$\therefore \int\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\int 2\cos x\ dx +\int 2\cos \alpha\ dx$
$=2[\sin x + x\cos \alpha]+C$

### Question:14  Find the integrals of the functions $\frac{\cos x - \sin x }{1+ \sin 2x }$

$\frac{\cos x-\sin x}{1+2\sin x}=\frac{\cos x-\sin x}{(sin^2x+cos^2x)+2 sin x.\cos x}$
$=\frac{\cos x-\sin x}{(\sin x+\cos x)^2}$

$\\sin x+\cos x =t\\ \therefore (\cos x-\sin x)dx = dt$

Now,
$=\int \frac{dt}{t^2}\\ =\int t^-2\ dt\\ =-t^{-1}+C\\ =-\frac{1}{(\sin x+\cos x)}+C$

### Question:15  Find the integrals of the functions  $\tan ^ 3 2x \sec 2x$

$\tan^32x.\sec 2x = \tan^22x.\tan 2x.\sec 2x$
$\\= (\sec^22x-1).\tan 2x.\sec 2x\\ =\sec^22x.\tan 2x-\tan 2x.\sec 2x$

Therefore integration of $\tan ^ 3 2x \sec 2x$ =
$\\=\int\sec^22x.\tan 2x\ dx-\int\tan 2x.\sec 2x\ dx\\ =\int\sec^22x.\tan 2x\ dx-\sec 2x/2+C\\$.....................(i)
Let assume

$\sec 2x = t$
So, that  $2\sec 2x.\tan 2x\ dx =dt$
Now, the equation (i) becomes,

$\\\Rightarrow \frac{1}{2}\int t^2\ dt-\frac{\sec 2x}{2}+C\\ \Rightarrow \frac{t^3}{6}-\frac{\sec 2x}{2}+C\\ =\frac{(\sec 2x)^3}{6}-\frac{\sec 2x}{2}+C$

### Question:16  Find the integrals of the functions  $\tan ^ 4x$

the given question can be rearranged using trigonometric identities

$tan^4x=(\sec^2x-1).\tan^2x\\ =\sec^2x.\tan^2x-\tan^2x\\ =\sec^2x.\tan^2x-\sec^2x+1$

Therefore, the integration of $\tan^4x$ = $\\=\int \sec^2x.\tan^2x\ dx-\int\sec^2x\ dx+\int dx\\ =(\int \sec^2x.\tan^2x\ dx)-\tan x+x+C\\$...................(i)
Considering only $\int \sec^2x.\tan^2x\ dx$
let $\tan x =t\Rightarrow \sec^2x\ dx =dt$

$\int \sec^2x\tan^2x\ dx = \int t^2\ dt = t^3/3=\frac{\tan^3x}{3}$

now the final solution is,

$\int \tan^4x =\frac{\tan^3x}{3}-\tan x+x+C$

### Question:17  Find the integrals of the functions $\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

$\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

now splitting the terms we can write

$\\=\frac{\sin^3x}{\sin^2x.\cos^2x}+\frac{\cos^3x}{\sin^2x.\cos^2x}\\ =\frac{\sin x}{cos^2x}+\frac{\cos x}{\sin^2x}\\ =\tan x.\sec x+\cot xcosec x$

Therefore, the integration of
$\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

$\\=\int (\tan x\sec x+\cot xcosec x)dx\\ =\sec x-cosec\ x+C$

### Question:18  Find the integrals of the functions $\frac{\cos 2 x + 2 \sin ^ 2x }{\cos ^ 2 x }$

The integral of the above equation is

$\\=\int (\frac{\cos 2x+2\sin^2x}{\cos^2x})dx\\ =\int (\frac{\cos 2x+(1-\cos 2x)}{\cos^2x}\\ =\int\frac{1}{\cos^2x}\\ =\int \sec^2x\ dx =\tan x+C$

Thus after evaluation, the value of integral is tanx+ c

### Question:19 Find the integrals of the functions $\frac{1}{\sin x \cos ^3 x }$

Let
We can write 1 = $\sin^2x +\cos^2x$
Then, the equation can be written as
$I =\frac{\sin^2x +\cos^2x}{\sin x\cos^3x}$

$I =\int (\tan x+\frac{1}{\tan x})\sec^2 x dx$
put the value of tan$x$ = t
So, that $\sec^2xdx =dt$

$\\\Rightarrow I=\int (t+\frac{1}{t})dt\\ =\frac{t^2}{2}+\log\left | t \right |+C\\ =\log\left | \tan x \right |+\frac{1}{2}\tan^2x+C$

### Question:20 Find the integrals of the functions $\frac{\cos 2x }{( \cos x + \sin x )^2}$

we know that $cos2x= cos^2x-sin^2x$
therefore,

$\frac{\cos 2x }{( \cos x + \sin x )^2}$
$\frac{\cos 2x}{1+\sin 2x}\\ \Rightarrow \int \frac{\cos 2x}{1+\sin 2x}\\$  let $1+sin2x =t \Rightarrow 2cos2x\ dx = dt$
Now the given integral can be written as

$\therefore \int \frac{\cos 2x}{(\cos x+\sin x)^2}=\frac{1}{2}\int \frac{1}{t}dt$
$\\\Rightarrow \frac{1}{2}\log\left | t \right |+C\\ \Rightarrow \frac{1}{2}\log\left | 1+\sin 2x \right |+C\\=log|sin^2x+cos^2x+2sinxcosx|+C\\=\frac{1}{2}log|(sinx+cosx)^2|+C=log|sinx+cosx|+C$

### Question:21  Find the integrals of the functions $\sin ^ { -1} ( \cos x )$

using the trigonometric identities we can evaluate the following integral as follows

$\dpi{100} \int \sin^{-1}(\cos x)dx = \int \sin^{-1}(sin(\frac{\pi}{2}-x))dx\\=\int(\frac{\pi}{2}-x)dx=\frac{\pi x}{2}-\frac{x^2}{2}+C$

### Question:22  Find the integrals of the functions $\frac{1}{\cos ( x-a ) \cos ( x-b )}$

Using the trigonometric identities following integrals can be simplified as follows

$\frac{1}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}]$

$=\frac{1}{\sin(a-b)}[\frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)}]$

$=\frac{1}{\sin(a-b)}[\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}]$

$=\frac{tan(x-b)-\tan (x-a)}{\sin(a-b)}$

$=\frac{1}{\sin(a-b)}\int tan(x-b)-\tan (x-a)dx$
$\\=\frac{1}{\sin(a-b)}[-\log\left | \cos(x-b) \right |+\log\left | \cos(x-a) \right |]\\ =\frac{1}{\sin(a-b)}(\log\left | \frac{\cos(x-a)}{\cos(x-b)} \right |)$

### Question:23 Choose the correct answer

$\int \frac{\sin ^ 2 x - \cos ^ 2 x dx }{\sin ^ 2 x \cos ^ 2x } dx \: \:is \: \:equal \: \: to$

$(A) \tan x + \cot x + C \\\\(B) \tan x + \cosec x + C\\\\ (C) - \tan x + \cot x + C\\\\ (D) \tan x + \sec x + C$

The correct option is (A)

On reducing the above integral becomes  $\sec^2x-csc^2x$
$\int\sec^2x-csc^2x\ dx = \tan x+ \cot x+C$

### Question:24 Choose the correct answer $\int \frac{e ^x ( 1+x)}{\cos ^ 2 ( e ^xx )} dx \: \: equals$

$\\(A) -\cot (ex^x) + C \\\\ (B) \tan (xe^x) + C\\\\ (C) \tan (e^x) + C \\\\ (D) \cot (e^x) + C$

The correct option is (B)

Let $e^xx = t$.
So, $(e^x.x+ 1.e^x)dx = dt$
(1+$x$)$e^x\ dx =dt$

therefore,

$\int \frac{e^x(1+x)}{\cos^2(e^x.x)}dx =\int\frac{dt}{\cos^2t}$
$\\=\int \sec^2t dt\\ =\tan t +C\\ =\tan(e^x.x)+C$

NCERT solutions for class 12 maths chapter 7 Integrals Exercise: 7.4

### Question:1 Integrate the functions $\frac{3x^ 2 }{x^6 + 1 }$

The given integral can be calculated as follows

Let $x^3 = t$
, therefore, $3x^2 dx =dt$

$\Rightarrow \int\frac{3x^2}{x^6+1}=\int \frac{dt}{t^2+1}$

$\\=\tan^{-1} t +C\\ =tan^{-1}(x^3)+C$

### Question:2  Integrate the functions $\frac{1}{\sqrt { 1+ 4 x^2 }}$

$\frac{1}{\sqrt { 1+ 4 x^2 }}$
let suppose 2x = t
therefore 2dx = dt

$\int \frac{1}{\sqrt{1+4x^2}} =\frac{1}{2}\int \frac{dt}{1+t^2}$
$\\=\frac{1}{2}[\log\left | t+\sqrt{1+t^2} \right |]+C\\ =\frac{1}{2}\log\left | 2x+\sqrt{4x^2+1} \right |+C$.................using formula $\dpi{100} \int\frac{1}{\sqrt{x^2+a^2}}dt = \log\left | x+\sqrt{x^2+a^2} \right |$

### Question:3  Integrate the functions $\frac{1}{\sqrt { ( 2- x)^2+ 1 }}$

$\frac{1}{\sqrt { ( 2- x)^2+ 1 }}$

let suppose 2-x =t
then, -dx =dt
$\Rightarrow \int\frac{1}{\sqrt{(2-x)^2+1}}dx = -\int \frac{1}{\sqrt{t^2+1}}dt$

using the identity

$\int \frac{1}{\sqrt{x^2+1}}dt=log\left | x+\sqrt{x^2+1} \right |$

$\\= -\log\left | t+\sqrt{t^2+1} \right |+C\\ =-\log\left | 2-x+\sqrt{(2-x)^2+1} \right |+C\\ =\log \left | \frac{1}{(2-x)+\sqrt{x^2-4x+5}} \right |+C$

### Question:4  Integrate the functions $\frac{1}{\sqrt {9 - 25 x^2 }}$

$\frac{1}{\sqrt {9 - 25 x^2 }}$
Let assume 5x =t,
then 5dx = dt

$\Rightarrow \int \frac{1}{\sqrt{9-25x^2}}=\frac{1}{5}\int \frac{1}{\sqrt{9-t^2}}dt$
$\\=\frac{1}{5}\int \frac{1}{\sqrt{3^2-t^2}}dt\\ =\frac{1}{5}\sin^{-1}(\frac{t}{3})+C\\ =\frac{1}{5}\sin^{-1}(\frac{5x}{3})+C$

The above result is obtained using the identity

$\\\int \frac{1}{\sqrt{a^2-x^2}}dt\\ =\frac{1}{a}sin^{-1}\frac{x}{a}$

### Question:5  Integrate the functions $\frac{3x }{1+ 2 x ^ 4 }$

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

Let ${\sqrt{2}}x^2 = t$
$\therefore$$2\sqrt{2}xdx =dt$

The integration can be done as follows

$\Rightarrow \int \frac{3x}{1+2x^4}= \frac{3}{2\sqrt{2}}\int \frac{dt}{1+t^2}$
$\\= \frac{3}{2\sqrt{2}}[\tan^{-1}t]+C\\ =\frac{3}{2\sqrt{2}}[\tan^{-1}(\sqrt{2}x^2)]+C$

### Question:6  Integrate the functions $\frac{x ^ 2 }{1- x ^ 6 }$

$\frac{x ^ 2 }{1- x ^ 6 }$

let $x^3 =t$
then $3x^2dx =dt$

using the special identities we can simplify the integral as follows

$\int \frac{x^2}{1-x^6}dx =\frac{1}{3}\int \frac{dt}{1-t^2}$
$=\frac{1}{3}[\frac{1}{2}\log\left | \frac{1+t}{1-t} \right |]+C\\ =\frac{1}{6}\log\left | \frac{1+x^3}{1-x^3} \right |+C$

### Question:7  Integrate the functions $\frac{x-1 }{\sqrt { x^2 -1 }}$

We can write above eq as
$\frac{x-1 }{\sqrt { x^2 -1 }}$$=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx$............................................(i)

for    $\int \frac{x}{\sqrt{x^2-1}}dx$                let $x^2-1 = t \Rightarrow 2xdx =dt$

$\therefore \int \frac{x}{\sqrt{x^2-1}}dx=\frac{1}{2}\int \frac{dt}{\sqrt{t}}$
$\\=\frac{1}{2}\int t^{1/2}dt\\ =\frac{1}{2}[2t^{1/2}]\\ =\sqrt{t}\\ =\sqrt{x^2-1}$
Now, by using eq (i)
$=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx$
$\\=\sqrt{x^2-1}-\int \frac{1}{\sqrt{x^2}-1}dx\\ =\sqrt{x^2-1}-\log\left | x+\sqrt{x^2-1} \right |+C$

### Question:8  Integrate the functions $\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}$

The integration can be down as follows

$\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}$
let $x^3 = t \Rightarrow 3x^2dx =dt$

$\therefore \frac{x^2}{\sqrt{x^6+a^6}}=\frac{1}{3}\int \frac{dt}{\sqrt{t^2+(a^3)^2}}$
$\\=\frac{1}{3}\log\left | t+\sqrt{t^2+a^6} \right |+C\\ =\frac{1}{3}\log\left | x^3+\sqrt{x^6+a^6} \right |+C$........................using $\int \frac{dx}{\sqrt{x^2+a^2}} = \log\left | x+\sqrt{x^2+a^2} \right |$

### Question:9 Integrate the functions $\frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x+ 4 }}$

The integral can be evaluated as follows

$\frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x + 4 }}$
let $\tan x =t \Rightarrow sec^2x dx =dt$

$\Rightarrow \int \frac{\sec^2x}{\sqrt{\tan^2x+4}}dx = \int \frac{dt}{\sqrt{t^2+2^2}}$
$\\= \log\left | t+\sqrt{t^2+4} \right |+C\\ =\log \left | \tan x+\sqrt{ tan^2x+4} \right |+C$

### Question:10 Integrate the functions $\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}$

$\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}$
the above equation can be also written as,
$=\int\frac{1}{\sqrt{(1+x)^2+1^2}}dx$
let 1+x = t
then dx = dt
therefore,

$\\=\int\frac{1}{\sqrt{t^2+1^2}}dx\\ =\log\left | t+\sqrt{t^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(1+x)^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(x^2+2x+2} \right |+C$

### Question:12  Integrate the functions $\frac{1}{\sqrt{ 7-6x - x ^ 2 }}$

the denominator can be also written as,
$7-6x-x^2=16-(x^2+6x+9)$
$=4^2-(x+3)^2$

therefore

$\int \frac{1}{\sqrt{7-6x-x^2}}dx=\int \frac{1}{\sqrt{4^2-(x+3)^2}}dx$
Let x+3 = t
then dx =dt

$\Rightarrow \int \frac{1}{\sqrt{4^2-(x+3)^2}}dx=\int \frac{1}{\sqrt{4^2-t^2}}dt$......................................using formula $\int \frac{1}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})$
$\\= sin^{-1}(\frac{t}{4})+C\\ =\sin^{-1}(\frac{x+3}{4})+C$

### Question:13  Integrate the functions$\frac{1}{\sqrt { ( x-1)( x-2 )}}$

(x-1)(x-2) can be also written as
$x^2-3x+2$
=$(x-\frac{3}{2})^2-(\frac{1}{2})^2$

therefore

$\int \frac{1}{\sqrt{(x-1)(x-2)}}dx= \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx$
let suppose
$x-3/2 = t \Rightarrow dx =dt$
Now,

$\Rightarrow \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx = \int \frac{1}{\sqrt{t^2-(\frac{1}{2})^2}}dt$.............by using formula $\int \frac{1}{\sqrt{x^2-a^2}}=\log\left | x+\sqrt{x^2+a^2} \right |$
$\\= \log \left | t+\sqrt{t^2-(1/2)^2} \right |+C\\ = \log \left | (x-\frac{3}{2})+\sqrt{x^2-3x+2} \right |+C$

### Question:14  Integrate the functions $\frac{1}{\sqrt { 8 + 3 x - x ^ 2 }}$

We can write denominator as
$\\=8-(x^2-3x+\frac{9}{4}-\frac{9}{4})\\ =\frac{41}{4}-(x-\frac{3}{2})^2$

therefore
$\Rightarrow \int \frac{1}{\sqrt{8+3x-x^2}}dx= \int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^2}}$
let $x-3/2 = t \Rightarrow dx =dt$

$\therefore$
$\\=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^2-t^2}}dt\\ =\sin^{-1}(\frac{t}{\frac{\sqrt{41}}{2}})+C\\ =\sin^{-1}(\frac{2x-3}{\sqrt{41}})+C$

### Question:15 Integrate the functions $\frac{1}{\sqrt {(x-a)( x-b )}}$

(x-a)(x-b) can be written as  $x^2-(a+b)x+ab$
$\\x^2-(a+b)x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}\\ (x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}$

$\Rightarrow \int\frac{1}{\sqrt{(x-a)(x-b)}}dx=\int \frac{1}{\sqrt{(x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}}}dx$
let
$x-\frac{(a+b)}{2}=t \Rightarrow dx =dt$
So,
$\\=\int \frac{1}{\sqrt{t^2-(\frac{a-b}{2})^2}}dt\\ =\log \left | t+\sqrt{t^2-(\frac{a-b}{2})^2} \right |+C\\ =\log \left | x-(\frac{a+b}{2})+\sqrt{(x-a)(x-b)} \right |+C$

### Question:16  Integrate the functions $\frac{4x+1 }{\sqrt {2x ^ 2 + x -3 }}$

let
$\\4x+1 = A\frac{d}{dx}(2x^2+x-3)+B\\ 4x+1=A(4x+1)+B\\ 4x+1=4Ax+A+B$

By equating the coefficient of x and constant term on each side, we get
A = 1 and B=0

Let $(2x^2+x-3) = t\Rightarrow (4x+1)dx =dt$

$\therefore \int \frac{4x+1}{\sqrt{2x^2+x-3}}dx= \int\frac{1}{\sqrt{t}}dt$
$\\= 2\sqrt{t}+C\\ =2\sqrt{2x^2+x-3}+C$

### Question:17 Integrate the functions $\frac{x+ 2 }{\sqrt { x ^2 -1 }}$

let $x+2 =A\frac{d}{dx}(x^2-1)+B=A(2x)+B$
By comparing the coefficients and constant term on both sides, we get;

A=1/2 and B=2
then  $x+2 = \frac{1}{2}(2x)+2$

$\int \frac{x+2}{\sqrt{x^2-1}}dx =\int\frac{1/2(2x)+2}{x^2-1}dx$
$\\=\frac{1}{2}\int\frac{(2x)}{\sqrt{x^2-1}}dx+\int \frac{2}{\sqrt{x^2-1}}dx\\ =\frac{1}{2}[2\sqrt{x^2-1}]+2\log\left | x+\sqrt{x^2-1} \right |+C\\ =\sqrt{x^2-1}+2\log\left | x+\sqrt{x^2-1} \right |+C$

### Question:18  Integrate the functions $\frac{5x -2 }{1+ 2x +3x^2 }$

let
$\\5x+2 = A\frac{d}{dx}(1+2x+3x^2)+B\\ 5x+2= A(2+6x)+B = 2A+B+6Ax$
By comparing the coefficients and constants we get the value of A and B

A = $5/6$ and B =$-11/3$

NOW,
$I = \frac{5}{6}\int \frac{6x+2}{3x^2+2x+1}dx-\frac{11}{3}\int \frac{dx}{3x^2+2x+1}$
$I = I_{1}-\frac{11}{3}I_{2}$...........................(i)

put $3x^2+2x+1 =t \Rightarrow (6x+2)dx =dt$
Thus
$I_{1}=\frac{5}{6}\int \frac{dt}{t} =\frac{5}{6}\log t =\frac{5}{6}\log (3x^2+2x+1)+c1$
$I_{2}= \int \frac{dx}{3x^2+2x+1} = \frac{1}{3}\int\frac{dx}{(x+1/3)^2+(\sqrt{2}/3)^2}$
$\\=\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt{2}})+c2$

$\therefore I = I_1+I_2$
$I = \frac{5}{6}\log(3x^2+2x+1)-\frac{11}{3}\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt2})+C$

### Question:19 Integrate the functions $\frac{6x + 7 }{\sqrt {( x-5 )( x-4)}}$

let
$6x+7 = A\frac{d}{dx}(x^2-9x+20)+B =A(2x-9)+B$
By comparing the coefficients and constants on both sides, we get
A =3  and B =34

$I =\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx = \int \frac{3(2x+9)}{\sqrt{x^2-9x+20}}dx+34\int\frac{dx}{\sqrt{x^2-9x+20}}$$I = I_1+I_2$....................................(i)

Considering $I_1$

$I_1 =\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx$        let $x^2-9x+20 = t \Rightarrow (2x-9)dx =dt$

$I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{x^2-9x+20}$

Now consider $I_2$

$I_2=\int \frac{dx}{\sqrt{x^2-9x+20}}$
here the denominator can be also written as
Dr = $(x-\frac{9}{2})^2-(\frac{1}{2})^2$

$\therefore I_2 = \int \frac{dx}{\sqrt{(x-\frac{9}{2})^2-(\frac{1}{2})^2}}$
$\\= \log\left | (x-\frac{9}{2})^2+\sqrt{x^2-9x+20} \right |$

Now put the values of $I_1$ and $I_2$ in eq (i)

$\\I = 3I_1+34I_2\\ I=6\sqrt{x^2-9x+20}+34\log\left | (x-\frac{9}{2})+\sqrt{x^2-9x+20} \right |+C$

### Question:20 Integrate the functions $\frac{x +2 }{\sqrt { 4x - x ^ 2 }}$

let
$x+2 = A\frac{d}{dx}(4x-x^2)+B = A(4-2x)+B$
By equating the coefficients and constant term on both sides  we get

A = -1/2 and B = 4

(x+2) = -1/2(4-2x)+4

$\\\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx = -\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}+4\int \frac{dx}{\sqrt{4x-x^2}}\\ \ I =\frac{-1}{2}I_1+4I_2$....................(i)

Considering $I_1$
$\int \frac{4-2x}{\sqrt{4x-x^2}}dx$
let $4x-x^2 =t \Rightarrow (4-2x)dx =dt$
$I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{4x-x^2}$
now, $I_2$

$I_2 =\int \frac{dx}{\sqrt{4x-x^2}} = \int \frac{dx}{\sqrt{2^2-(x-2)^2}}$
$=\sin^{-1}(\frac{x-2}{2})$

put the value of $I_1$and $I_2$

$I =-\sqrt{4x-x^2}+4\sin^{-1}(\frac{x-2}{2})+C$

### Question:21 Integrate the functions $\frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}$

$\frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}$
$\int \frac{x+2}{\sqrt{x^2+2x+3}}dx = \frac{1}{2}\int \frac{2(x+2)}{\sqrt{x^2+2x+3}}dx$
$\\= \frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\frac{1}{2}\int \frac{2}{\sqrt{x^2+2x+3}}dx\\ =\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\int \frac{1}{\sqrt{x^2+2x+3}}dx\\ I=\frac{1}{2}I_1+I_2$...........(i)

take $I_1$

$\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx$
let $x^2+2x+3 = t \Rightarrow (2x+2)dx =dt$

$I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+2x+3}$
considering $I_2$

$= \int \frac{dx}{\sqrt{x^2+2x+3}}= \int \frac{dx}{\sqrt{(x+1)^2+(\sqrt{2})^2}}$
$= \log \left | (x+1)+\sqrt{x^2+2x+3} \right |$
putting the values  in equation (i)

$I=\sqrt{x^2+2x+3} +\log \left | (x+1)+\sqrt{x^2+2x+3} \right |+C$

### Question:22 Integrate the functions $\frac{x + 3 }{x ^ 2 - 2x - 5 }$

Let $(x+3) =A\frac{d}{dx}(x^2-2x+5)+B= A(2x-2)+B$

By comparing the coefficients and constant term, we get;

A = 1/2 and B =4

$\\\int \frac{x+3}{x^2-2x+5}dx = \frac{1}{2}\int \frac{2x-2}{x^2-2x+5}dx +4\int \frac{1}{x^2-2x+5}dx\\ I=I_1+I_2$..............(i)

$\\\Rightarrow I_1\\ =\int \frac{2x-2}{x^2-2x-5}dx$
put $x^2-2x-5 =t \Rightarrow (2x-2)dx =dt$

$=\int \frac{dt}{t} = \log t = \log (x^2-2x-5)$

$\\\Rightarrow I_2\\ = \int \frac{1}{x^2-2x-5}dx\\ =\int \frac{1}{(x-1)^2+(\sqrt{6})^2}dx\\ =\frac{1}{2\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})$

$I=I_1+I_2$

$=\frac{1}{2}\log\left | x^2-2x-5 \right |+\frac{2}{\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})+C$

### Question:23 Integrate the functions $\frac{5x + 3 }{\sqrt { x^2 + 4x +10 }}$

let
$5x+3 = A\frac{d}{dx}(x^2+4x+10)+B = A(2x+4)+B$
On comparing, we get

A =5/2 and B = -7

$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx = \frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}dx$$I = 5/2I_1-7I_2$...........................................(i)

$\\\Rightarrow I_1\\ \int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$
put
$x^2+4x+10= t \Rightarrow (2x+4)dx = dt$

$=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+4x+10}$

$\\\Rightarrow I_2\\ =\int \frac{1}{\sqrt{x^2+4x+10}}dx \\ =\int \frac{1}{\sqrt{(x+2)^2+(\sqrt{6})^2}}dx\\ =\log \left | (x+2)+\sqrt{x^2+4x+10} \right |$

$I = 5\sqrt{x^2+4x+10}-7\log\left | (x+2)+\sqrt{x^2+4x+10} \right |+C$

### Question:24  Choose the correct answer

$\int \frac{dx }{x^2 + 2x +2 }\: \: equals$

$(A) x \tan^{-1} (x + 1) + C\\\\ (B) \tan^{-1} (x + 1) + C\\\\ (C) (x + 1) \tan^{-1}x + C \\\\ (D) \tan^{-1}x + C$

The correct option is (B)

$\int \frac{dx }{x^2 + 2x +2 }\: \: equals$
the denominator can be written as  $(x+1)^2+1$
now, $\int \frac{dx}{(x+1)^2+1} = tan^{-1}(x+1)+C$

### Question:25 Choose the correct answer $\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals$

$A) \frac{1}{9} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\B ) \frac{1}{2} \sin ^{-1}\left ( \frac{8x-9}{9} \right )+ C \\\\ C) \frac{1}{3} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\ D ) \frac{1}{2} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C$

The following integration can be done as

$\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals$
$\int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x)}}= \int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x+81/64-81/64)}}dx$
$\\= \int \frac{1}{\sqrt{-4[(x-9/8)^2-(9/8)^2]}}dx\\ =\frac{1}{2}\int \frac{1}{\sqrt{-(x-9/8)^2+(9/8)^2}}dx\\ =\frac{1}{2}[\sin^{-1}(\frac{x-9/8}{9/8})]+C\\ =\frac{1}{2}\sin^{-1}(\frac{8x-9}{9})+C$

The correct option is (B)

NCERT solutions for class 12 maths chapter 7 Integrals Exercise: 7.5

Given function $\frac{x }{( x +1)( x+2)}$

Partial function of this function:

$\frac{x }{( x +1)( x+2)} = \frac{A}{(x+1)}+\frac{B}{(x+2)}$

$\implies x = A(x+2)+B(x+1)$

Now, equating the coefficients of x and constant term, we obtain

$A+B =1$

$2A+B =0$

On solving, we get

$A=-1\ and\ B =2$

$\therefore \frac{x}{(x+1)(x+2)} = \frac{-1}{(x+1)}+\frac{2}{(x+2)}$

$\implies \int \frac{x}{(x+1)(x+2)} dx =\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} dx$

$=-\log|x+1| +2\log|x+2| +C$

$=\log(x+2)^2-\log|x+1|+C$

$=\log\frac{(x+2)^2}{(x+1)}+C$

### Question:2  Integrate the rational functions $\frac{1}{x^2 -9 }$

Given function $\frac{1}{x^2 -9 }$

The partial function of this function:

$\frac{1}{(x+3)(x-3)}= \frac{A}{(x+3)}+\frac{B}{(x-3)}$

$1 = A(x-3)+B(x+3)$

Now, equating the coefficients of x and constant term, we obtain

$A+B =1$

$-3A+3B =1$

On solving, we get

$A=-\frac{1}{6}\ and\ B =\frac{1}{6}$

$\frac{1}{(x+3)(x-3)}= \frac{-1}{6(x+3)} +\frac{1}{6(x-3)}$

$\int \frac{1}{(x^2-9)}dx = \int \left ( \frac{-1}{6(x+3)}+\frac{1}{6(x-3)} \right )dx$

$=-\frac{1}{6}\log|x+3| +\frac{1}{6}\log|x-3| +C$

$= \frac{1}{6}\log\left | \frac{x-3}{x+3} \right |+C$

### Question:3 Integrate the rational functions $\frac{3x -1}{( x-1)(x-2)(x-3)}$

Given function $\frac{3x -1}{( x-1)(x-2)(x-3)}$

Partial function of this function:

$\frac{3x -1}{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ .(1)

Now, substituting $x=1,2,\ and\ 3$ respectively in equation (1), we get

$A =1,\ B=-5,\ and\ C=4$

$\therefore \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{(x-1)} -\frac{5}{(x-2)}+\frac{4}{(x-3)}$

That implies $\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)} \right \}dx$

$= \log|x-1|-5\log|x-2|+4\log|x-3|+C$

### Question:4  Integrate the rational functions $\frac{x }{( x-1)(x-2)(x-3)}$

Given function $\frac{x }{( x-1)(x-2)(x-3)}$

Partial function of this function:

$\frac{x }{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$x = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ .....(1)

Now, substituting $x=1,2,\ and\ 3$ respectively in equation (1), we get

$A =\frac{1}{2},\ B=-2,\ and\ C=\frac{3}{2}$

$\therefore \frac{x}{(x-1)(x-2)(x-3)} = \frac{1}{2(x-1)} -\frac{2}{(x-2)}+\frac{3}{2(x-3)}$

That implies $\int \frac{x}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)} \right \}dx$

$= \frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C$

### Question:5  Integrate the rational functions $\frac{2x}{x^2 + 3x +2 }$

Given function $\frac{2x}{x^2 + 3x +2 }$

Partial function of this function:

$\frac{2x}{x^2 + 3x +2 }= \frac{A}{(x+1)}+\frac{B}{(x+2)}$

$2x = A(x+2)+B(x+1)$                                  ...........(1)

Now, substituting $x=-1\ and\ -2$ respectively in equation (1), we get

$A ={-2},\ B=4$

$\frac{2x}{x^2 + 3x +2 }= \frac{-2}{(x+1)}+\frac{4}{(x+2)}$

That implies $\int \frac{2x}{x^2 + 3x +2 }dx= \int \left \{ \frac{-2}{(x+1)}+\frac{4}{(x+2)} \right \}dx$

$=4\log|x+2| -2\log|x+1| +C$

### Question:6  Integrate the rational functions $\frac{1- x^2 }{ x ( 1- 2x )}$

Given function $\frac{1- x^2 }{ x ( 1- 2x )}$

Integral is not a proper fraction so,

Therefore, on dividing $(1-x^2)$ by $x(1-2x)$, we get

$\frac{1- x^2 }{ x ( 1- 2x )} = \frac{1}{2} +\frac{1}{2}\left ( \frac{2-x}{x(1-2x)} \right )$

Partial function of this function:

$\frac{2-x}{x(1-2x)} =\frac{A}{x}+\frac{B}{(1-2x)}$

$(2-x) =A(1-2x)+Bx$                                  ...........(1)

Now, substituting $x=0\ and\ \frac{1}{2}$  respectively in equation (1), we get

$A =2,\ B=3$

$\therefore \frac{2-x}{x(1-2x)} = \frac{2}{x}+\frac{3}{1-2x}$

No, substituting in equation (1) we get

$\frac{1-x^2}{(1-2x)} = \frac{1}{2}+\frac{1}{2}\left \{ \frac{2}{3}+\frac{3}{(1-2x)} \right \}$

$\implies \int \frac{1-x^2}{x(1-2x)}dx =\int \left \{ \frac{1}{2}+\frac{1}{2}\left ( \frac{2}{x}+\frac{3}{1-2x} \right ) \right \}dx$

$=\frac{x}{2}+\log|x| +\frac{3}{2(-2)}\log|1-2x| +C$

$=\frac{x}{2}+\log|x| -\frac{3}{4}\log|1-2x| +C$

### Question:7 Integrate the rational functions $\frac{x }{( x^2+1 )( x-1)}$

Given function $\frac{x }{( x^2+1 )( x-1)}$

Partial function of this function:

$\frac{x }{( x^2+1 )( x-1)} = \frac{Ax+b}{(x^2+1)} +\frac{C}{(x-1)}$

$x = (Ax+B)(x-1)+C(x^2+1)$

$x=Ax^-Ax+Bc-B+Cx^2+C$

Now, equating the coefficients of $x^2, x$ and the constant term, we get

$A+C = 0$

$-A+B =1$   and  $-B+C = 0$

On solving these equations, we get

$A = -\frac{1}{2}, B= \frac{1}{2},\ and\ C=\frac{1}{2}$

From equation (1), we get

$\therefore \frac{x}{(x^2+1)(x-1)} = \frac{\left ( -\frac{1}{2}x+\frac{1}{2} \right )}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}$

$\implies \int \frac{x}{(x^2+1)(x-1)}$

$=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2} \int \frac{1}{x-1}dx$

$=- \frac{1}{4} \int \frac{2x}{x^2+1} dx +\frac{1}{2} \tan^{-1}x + \frac{1}{2} \log|x-1| +C$

Now, consider  $\int \frac{2x}{x^2+1} dx$

and we will assume $(x^2+1) = t \Rightarrow 2xdx =dt$

So, $\int \frac{2x}{x^2+1}dx = \int \frac{dt}{t} =\log|t| = \log|x^2+1|$

$\therefore \int \frac{x}{(x^2+1)(x-1)} =-\frac{1}{4}\log|x^2+1| +\frac{1}{2}\tan^{-1}x +\frac{1}{2}\log|x-1| +C$or

$\frac{1}{2}\log|x-1| - \frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x +C$

### Question:8 Integrate the rational functions $\frac{x }{( x+1)^2 ( x+2)}$

Given function $\frac{x }{( x+1)^2 ( x+2)}$

Partial function of this function:

$\frac{x }{( x+1)^2 ( x+2)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}$

$x = A(x-1)(x+2)+B(x+2)+C(x-1)^2$

Now, putting $x=1$ in the above equation, we get

$B =\frac{1}{3}$

By equating the coefficients of $x^2$ and constant term, we get

$A+C=0$

$-2A+2B+C = 0$

then after solving, we get

$A= \frac{2}{9}\ and\ C=\frac{-2}{9}$

Therefore,

$\frac{x}{(x-1)^2(x+2)} = \frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}$

$\int \frac{x}{(x-1)^2(x+2)}dx= \frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1)^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx$

$= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C$

$\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C$

### Question:9  Integrate the rational functions $\frac{3x+ 5 }{x^3 - x^2 - x +1 }$

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

can be rewritten as $\frac{3x+ 5 }{x^3 - x^2 - x +1 } = \frac{3x+5}{(x-1)^2(x+1)}$

Partial function of this function:

$\frac{3x+5}{(x-1)^2(x+1)}= \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}$

$3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)^2$

$3x+5 = A(x^2-1)+B(x+1)+C(x^2+1-2x)$      ................(1)

Now, putting $x=1$ in the above equation, we get

$B =4$

By equating the coefficients of $x^2$ and $x$ , we get

$A+C=0$

$B-2C =3$

then after solving, we get

$A= -\frac{1}{2}\ and\ C=\frac{1}{2}$

Therefore,

$\frac{3x+5}{(x-1)^2(x+1)}= \frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}$

$\int \frac{3x+5}{(x-1)^2(x+1)}dx= \frac{-1}{2}\int \frac{1}{(x-1)}dx+4\int \frac{1}{(x-1)^2} dx+\frac{1}{2}\int \frac{1}{(x+1)}dx$

$= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C$

$=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C$

### Question:10 Integrate the rational functions $\frac{2x -3 }{(x^2 -1 )( 2x+3)}$

Given function $\frac{2x -3 }{(x^2 -1 )( 2x+3)}$

can be rewritten as $\frac{2x -3 }{(x^2 -1 )( 2x+3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}$

The partial function of this function:

$\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{A}{(x+1)} +\frac{B}{(x-1)}+\frac{C}{(2x+3)}$

$\Rightarrow (2x-3) =A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)$ $\Rightarrow (2x-3) =A(2x^2+x-3)+B(2x^2+5x+3)+C(x^2-1)$$\Rightarrow (2x-3) =(2A+2B+C)x^2+(A+5B)x+(-3A+3B-C)$

Equating the coefficients of $x^2\ and\ x$, we get

$B=-\frac{1}{10},\ A =\frac{5}{2},\ and\ C= -\frac{24}{5}$

Therefore,

$\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{5}{2(x+1)} -\frac{1}{10(x-1)}-\frac{24}{5(2x+3)}$

$\implies \int \frac{2x-3}{(x^2-1)(2x+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}{(2x+3)}dx$$= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{24}{10}\log|2x+3|$

$= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{12}{5}\log|2x+3|+C$

$= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C$

$=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C$

$= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C$

$\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C$

### Question:11 Integrate the rational functions $\frac{5x}{(x+1)(x^2-4)}$

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

can be rewritten as $\frac{5x}{(x+1)(x^2-4)} = \frac{5x}{(x+1)(x+2)(x-2)}$

The partial function of this function:

$\frac{5x }{(x+1)( x+2)(x-2)} = \frac{A}{(x+1)} +\frac{B}{(x+2)}+\frac{C}{(x-2)}$

$\Rightarrow (5x) =A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)$

Now, substituting the value of $x =-1,-2,\ and\ 2$ respectively in the equation above, we get

$A=\frac{5}{3},\ B =\frac{-5}{2},\ and\ C= \frac{5}{6}$

Therefore,

$\frac{5x }{(x+1)( x+2)(x-2)} = \frac{5}{3(x+1)} -\frac{5}{2(x+2)}+\frac{5}{6(x-2)}$

$\implies \int \frac{5x}{(x+1)(x^2-4)}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|+C$

### Question:12 Integrate the rational functions $\frac{x^3 + x +1}{ x^2-1}$

Given function $\frac{x^3 + x +1}{ x^2-1}$

As the given integral is not a proper fraction.

So, we divide $(x^3+x+1)$ by $x^2-1$, we get

$\frac{x^3 + x +1}{ x^2-1} = x+\frac{2x+1}{x^2-1}$

can be rewritten as $\frac{2x+1}{x^2-1} =\frac{A}{(x+1)} +\frac{B}{(x-1)}$

$2x+1 ={A}{(x-1)} +{B}{(x+1)}$                           ....................(1)

Now, substituting $x =1\ and\ x=-1$ in equation (1), we get

$A =\frac{1}{2}\ and\ B=\frac{3}{2}$

Therefore,

$\frac{x^3+x+1 }{(x^2-1)} =x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$

$\implies \int \frac{x^3+x+1 }{(x^2-1)}dx =\int xdx +\frac{1}{2}\int \frac{1}{(x+1)} dx+\frac{3}{2} \int \frac{1}{(x-1)}dx$

$= \frac{x^2}{2}+\frac{1}{2}\log|x+1| +\frac{3}{2}\log|x-1| +C$

### Question:13 Integrate the rational functions $\frac{2}{(1-x)(1+ x^2)}$

Given function $\frac{2}{(1-x)(1+ x^2)}$

can be rewritten as $\frac{2}{(1-x)(1+ x^2)} = \frac{A}{(1-x)}+\frac{Bx+C}{1+x^2}$

$2 =A(1+x^2)+(Bx+C)(1-x)$                           ....................(1)

$2 =A +Ax^2 +Bx-Bx^2+C-Cx$

Now, equating the coefficient of $x^2, x,$ and constant term, we get

$A-B= 0$$B-C = 0$, and $A+C =2$

Solving these equations, we get

$A=1, B=1,\ and\ C=1$

Therefore,

$\therefore \frac{2}{(1-x)(1+ x^2)} = \frac{1}{(1-x)}+\frac{x+1}{1+x^2}$

$\implies \int \frac{2}{(1-x)(1+ x^2)}dx =\int \frac{1}{(1-x)} dx+ \int \frac{x}{1+x^2}dx +\int \frac{1}{1+x^2}dx$$= -\int \frac{1}{x-1}dx +\frac{1}{2}\int \frac{2x}{1+x^2}dx +\int\frac{1}{1+x^2}dx$

$=-\log|x-1| +\frac{1}{2}\log|1+x^2| +\tan^{-1}x+C$

### Question:14 Integrate the rational functions $\frac{3x-1}{(x+2)^2}$

Given function $\frac{3x-1}{(x+2)^2}$

can be rewritten as $\frac{3x-1}{(x+2)^2} = \frac{A}{(x+2)}+\frac{B}{(x+2)^2}$

$3x-1 = A(x+2)+B$

Now, equating the coefficient of $x$ and constant term, we get

$A=3$ and   $2A+B = -1$

Solving these equations, we get

$B=-7$

Therefore,

$\frac{3x-1}{(x+2)^2} = \frac{3}{(x+2)}-\frac{7}{(x+2)^2}$

$\implies \int\frac{3x-1}{(x+2)^2}dx = 3 \int \frac{1}{(x+2)}dx-7\int \frac{x}{(x+2)^2}dx$

$\implies 3\log|x+2| -7\left ( \frac{-1}{(x+2)}\right )+C$

$\implies 3\log|x+2| + \frac{7}{(x+2)} +C$

### Question:15 Integrate the rational functions $\frac{1}{x^4 -1 }$

Given function $\frac{1}{x^4 -1 }$

can be rewritten as $\frac{1}{x^4 -1 } = \frac{1}{(x^2-1)(x^2+1)} =\frac{1}{(x+1)(x-1)(1+x^2)}$

The partial fraction of above equation,

$\frac{1}{(x+1)(x-1)(1+x^2)} = \frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{Cx+D}{(x^2+1)}$

$1 = A(x-1)(x^2+1) +B(x+1)(x^2+1)+(Cx+D)(x^2-1)$

$1 = A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+Cx^3+Dx^2-Cx-D$$1 = (A+B+C)x^3 +(-A+B+D)x^2+(A+B-C)x+(-A+B-D)$

Now, equating the coefficient of $x^3,x^2,x$ and constant term, we get

$A+B+C = 0$ and   $-A+B+D = 0$

$A+B-C = 0$   and   $-A+B-D = 1$

Solving these equations, we get

$A= -\frac{1}{4}, B=\frac{1}{4},C=0,\ and\ D = -\frac{1}{2}$

Therefore,

$\frac{1}{x^4-1} = \frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^2+1)}$

$\implies \int \frac{1}{x^4-1}dx = -\frac{1}{4}\log|x-1| +\frac{1}{4}\log|x-1| -\frac{1}{2}\tan^{-1}x +C$

$= \frac{1}{4}\log|\frac{x-1}{x+1}| -\frac{1}{2}\tan^{-1}x +C$

### Question:16 Integrate the rational functions $\frac{1}{x ( x^n+1)}$

[Hint: multiply numerator and denominator by $x ^{n-1}$ and put $x ^n = t$ ]

Given function $\frac{1}{x ( x^n+1)}$

Applying Hint multiplying numerator and denominator by $x^{n-1}$ and putting $x^n =t$

$\frac{1}{x ( x^n+1)} = \frac{x^{n-1}}{x^{n-1}x(x^n+1)} = \frac{x^{n-1}}{x^n(x^n+1)}$

Putting $x^n =t$

$\therefore x^{n-1}dx =dt$

can be rewritten as $\int \frac{1}{x ( x^n+1)}dx =\int \frac{x^{n-1}}{x^n(x^n+1)}dx = \frac{1}{n} \int \frac{1}{t(t+1)}dt$

Partial fraction of above equation,

$\frac{1}{t(t+1)} =\frac{A}{t}+\frac{B}{(t+1)}$

$1 = A(1+t)+Bt$                                      ................(1)

Now, substituting $t = 0,-1$ in equation (1), we get

$A=1\ and\ B=-1$

$\therefore \frac{1}{t(t+1)} = \frac{1}{t}- \frac{1}{(1+t)}$

$\implies \int \frac{1}{x(x^n+1)}dx = \frac{1}{n} \int \left \{ \frac{1}{t}-\frac{1}{(t+1)} \right \}dx$

$= \frac{1}{n} \left [ \log|t| -\log|t+1| \right ] +C$

$= -\frac{1}{n} \left [ \log|x^n| -\log|x^n+1| \right ] +C$

$= \frac{1}{n} \log|\frac{x^n}{x^n+1}| +C$

### Question:17 Integrate the rational functions $\frac{\cos x }{(1- \sin x )( 2- \sin x )}$

[Hint : Put $\sin x = t$]

Given function $\frac{\cos x }{(1- \sin x )( 2- \sin x )}$

Applying the given hint: putting $\sin x =t$

We get, $\cos x dx =dt$

$\therefore \int \frac{\cos x }{(1- \sin x )( 2- \sin x )}dx = \int \frac{dt}{(1-t)(2-t)}$

Partial fraction of above equation,

$\frac{1}{(1-t)(2-t)} =\frac{A}{(1-t)}+\frac{B}{(2-t)}$

$1 = A(2-t)+B(1-t)$                                      ................(1)

Now, substituting $t = 2\ and\ 1$ in equation (1), we get

$A=1\ and\ B=-1$

$\therefore \frac{1}{(1-t)(2-t)} = \frac{1}{(1-t)} - \frac{1}{(2-t)}$

$\implies \int \frac{\cos x }{(1-\sin x)(2-\sin x )}dx = \int \left \{ \frac{1}{1-t}-\frac{1}{(2-t)} \right \}dt$

$= -\log|1-t| +\log|2-t| +C$

$= \log\left | \frac{2-t}{1-t} \right |+C$

Back substituting the value of t in the above equation, we get

$= \log\left | \frac{2-\sin x}{1- \sin x} \right |+C$

### Question:18 Integrate the rational functions $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}$

Given function $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}$

We can rewrite it as: $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \frac{(4x^2+10)}{(x^2+3)(x^2+4)}$

Partial fraction of above equation,

$\frac{(4x^2+10)}{(x^2+3)(x^2+4)} =\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}$

$4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)$

$4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D$

$4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(3D+4B)$

Now, equating the coefficients of $x^3, x^2, x$ and constant term, we get

$A+C=0$$B+D = 4$$4A+3C = 0$$4B+3D =10$

After solving these equations, we get

$A= 0, B =-2, C=0,\and\ D=6$

$\therefore \frac{4x^2+10}{(x^2+3)(x^2+4)} = \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)}$

$\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \left ( \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)} \right )$

$\implies \int \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} dx= \int \left \{ 1+ \frac{2}{(x^2+3)} - \frac{6}{(x^2+4)} \right \}dx$

$= \int \left \{ 1+ \frac{2}{(x^2+(\sqrt3)^2)} - \frac{6}{(x^2+2^2)} \right \}dx$

$= x+2\left ( \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt 3} \right ) - 6\left ( \frac{1}{2}\tan^{-1}\frac{x}{2} \right )+C$

$= x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} -3\tan^{-1}\frac{x}{2}+C$

### Question:19 Integrate the rational functions $\frac{2x }{( x^2 +1)( x^2 +3)}$

Given function $\frac{2x }{( x^2 +1)( x^2 +3)}$

Taking $x^2 = t \Rightarrow 2xdx=dt$

$\therefore \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \frac{dt}{(t+1)(t+3)}$

The partial fraction of above equation,

$\frac{1}{(t+3)(t+3)} = \frac{A}{(t+1)}+\frac{B}{(t+3)}$

$1= A(t+3)+B(t+1)$                                            ..............(1)

Now, substituting $t = -3\ and\ t = -1$ in equation (1), we get

$A =\frac{1}{2}\ and\ B = -\frac{1}{2}$

$\therefore\frac{1}{(t+3)(t+3)} = \frac{1}{2(t+1)}-\frac{1}{2(t+3)}$

$\implies \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \left \{ \frac{1}{2(t+1)}-\frac{1}{2(t+3)} \right \}dt$

$= \frac{1}{2}\log|t+1|- \frac{1}{2}\log|t+3| +C$

$= \frac{1}{2}\log\left | \frac{t+1}{t+3} \right | +C$

$= \frac{1}{2}\log\left | \frac{x^2+1}{x^2+3} \right | +C$

### Question:20 Integrate the rational functions $\frac{1}{x (x^4 -1)}$

Given function $\frac{1}{x (x^4 -1)}$

So, we multiply numerator and denominator by $x^3$, to obtain

$\frac{1}{x (x^4 -1)} = \frac{x^3}{x^4(x^4-1)}$

$\therefore \int \frac{1}{x(x^4-1)}dx =\int\frac{x^3}{x^4(x^4-1)}dx$

Now, putting $x^4 = t$

we get, $4x^3dx =dt$

Taking $x^2 = t \Rightarrow 2xdx=dt$

$\therefore \int \frac{1}{x(x^4-1)}dx =\frac{1}{4}\int \frac{dt}{t(t-1)}$

Partial fraction of above equation,

$\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}$

$1= A(t-1)+Bt$                                            ..............(1)

Now, substituting $t = 0\ and\ t = 1$ in equation (1), we get

$A = -1\ and\ B=1$

$\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}$

$\Rightarrow \int \frac{1}{x(x^4+1)}dx =\frac{1}{4}\int \left \{ \frac{-1}{t}+\frac{1}{t-1} \right \}dt$

$= \frac{1}{4} \left [ -\log|t|+\log|t-1| \right ]+C$

$= \frac{1}{4}\log\left | \frac{t-1}{t} \right |+C$

Back substituting the value of t,

$=\frac{1}{4}\log \left | \frac{x^4-1}{x^4} \right | +C$

### Question:21 Integrate the rational functions $\frac{1}{( e ^x-1)}$ [Hint : Put $e ^x= t$]

Given function $\frac{1}{( e ^x-1)}$

So, applying the hint: Putting $e^x = t$

Then $e^x dx= dt$

$\int \frac{1}{( e ^x-1)}dx = \int\frac{1}{t-1}\times\frac{dt}{t} = \int \frac{1}{t(t-1)}dt$

Partial fraction of above equation,

$\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}$

$1= A(t-1)+Bt$                                            ..............(1)

Now, substituting $t = 0\ and\ t = 1$ in equation (1), we get