# NCERT Solutions for Exercise 7.6 Class 12 Maths Chapter 7 - Integrals

In NCERT solutions for Class 12 Maths chapter 7 exercise 7.6 functions like logarithmic, inverse trigonometric functions, exponential functions etc. are discussed in step by step manner. Solutions to exercise 7.6 Class 12 Maths are prepared in a holistic manner by subject matter experts. Hence NCERT solutions for Class 12 Maths chapter 7 exercise 7.6 provided below can be a good source. There is no need to refer to any other book for these topics as the content provided in the NCERT book is more than sufficient for the examination. Practicing from this exercise can help students score well in exams like JEE Main. The NCERT chapter Integrals also has the following exercise for practice.

• Integrals Exercise 7.1

• Integrals Exercise 7.2

• Integrals Exercise 7.3

• Integrals Exercise 7.4

• Integrals Exercise 7.5

• Integrals Exercise 7.7

• Integrals Exercise 7.8

• Integrals Exercise 7.9

• Integrals Exercise 7.10

• Integrals Exercise 7.11

• Integrals Miscellaneous Exercise

Integrals Class 12 Chapter 7 Exercise: 7.6

Question:1 Integrate the functions $x \sin x$

Given function is
$f(x)=x \sin x$
We will use integrate by parts method
$\int x\sin x = x.\int \sin xdx - \int(\frac{d(x)}{dx}.\int sin x dx)dx\\ \\ \int x\sin x = x.(-\cos x)- \int (1.(-\cos x))dx\\ \\ \int x\sin x= -x\cos x+\sin x + C$
Therefore, the answer is $-x\cos x+\sin x + C$

### Question:2 Integrate the functions $x \sin 3x$

Given function is
$f(x)=x \sin 3x$
We will use integration by parts method
$\int x\sin 3x = x.\int \sin 3xdx - \int(\frac{d(x)}{dx}.\int sin 3x dx)dx\\ \\ \int x\sin 3x = x.(\frac{-\cos 3x}{3})- \int (1.(\frac{-\cos 3x}{3}))dx\\ \\ \int x\sin 3x= -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C$

Therefore, the answer is $-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C$

### Question:3 Integrate the functions $x ^ 2 e ^x$

Given function is
$f(x)=x^2e^x$
We will use integration by parts method
$\int x^2e^x= x^2.\int e^xdx - \int(\frac{d(x^2)}{dx}.\int e^x dx)dx\\ \\ \int x^2e^x = x^2.e^x- \int (2x.e^x)dx\\$
Again use integration by parts in $\int (2x.e^x)dx\\$
$\int (2x.e^x)dx = 2x.\int e^x dx - \int (\frac{d(2x)}{dx}.\int e^xdx)dx\\ \int 2x.e^x dx = 2xe^x- \int 2.e^xdx\\ \int 2x.e^x dx = 2xe^x- 2e^x$
Put this value in our equation
we will get,
$\int x^2.e^x dx =x^2e^x -2xe^x+ 2e^x + C\\ \int x^2.e^x dx = e^x(x^2-2x+2)+ C$

Therefore, answer is $e^x(x^2-2x+2)+ C$

### Question:4 Integrate the functions $x \log x$

Given function is
$f(x)=x.\log x$
We will use integration by parts method
$\int x.\log xdx= \log x.\int xdx - \int(\frac{d(\log x)}{dx}.\int x dx)dx\\ \\ \int x\log xdx = \log x.\frac{x^2}{2}- \int (\frac{1}{x}.\frac{x^2}{2})dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

Therefore, the answer is $\frac{x^2}{2}\log x- \frac{x^2}{4}+ C$

### Question:5 Integrate the functions $x \log 2x$

Given function is
$f(x)=x.\log 2 x$
We will use integration by parts method
$\int x.\log 2xdx= \log 2x.\int xdx - \int(\frac{d(\log 2x)}{dx}.\int x dx)dx\\ \\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int (\frac{2}{2x}.\frac{x^2}{2})dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

Therefore, the answer is $\log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

### Question:6 Integrate the functions $x^ 2 \log x$

Given function is
$f(x)=x^2.\log x$
We will use integration by parts method
$\int x^2.\log xdx= \log x.\int x^2dx - \int(\frac{d(\log x)}{dx}.\int x^2 dx)dx\\ \\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int (\frac{1}{x}.\frac{x^3}{3})dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int \frac{x^2}{3}dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \frac{x^3}{9}+ C$

Therefore, the answer is $\log x.\frac{x^3}{3}- \frac{x^3}{9}+ C$

### Question:7 Integrate the functions $x \sin ^{ -1} x$

Given function is
$f(x)=x.\sin^{-1} x$
We will use integration by parts method
$\int x.\sin^{-1} xdx= \sin^{-1} x.\int xdx - \int(\frac{d(\sin^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
Now, we need to integrate $\int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x-\sin^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\sin^{-1}x}{4}+C$
Put this value in our equation

Therefore, the answer is $\int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}$

### Question:8 Integrate the functions $x \tan ^{-1} x$

Given function is
$f(x)=x.\tan^{-1} x$
We will use integration by parts method
$\int x.\tan^{-1} xdx= \tan^{-1} x.\int xdx - \int(\frac{d(\tan^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}- \int (\frac{1}{1+x^2}.\frac{x^2}{2})dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( \frac{x^2+1}{1+x^2}-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( 1-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\left ( x- \tan^{-1}x \right )+C\\ \\ \int x\tan^{-1}xdx = \frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$

Put this value in our equation
$\int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \frac{x}{4\sqrt{1-x^2}}-\frac{\sin^{-1}x}{4}+C\\ \int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x}{4\sqrt{1-x^2}}$

Therefore, the answer is $\frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$

### Question:9 Integrate the functions $x\cos ^{ -1} x$

Given function is
$f(x)=x.\cos^{-1} x$
We will use integration by parts method
$\int x.\cos^{-1} xdx= \cos^{-1} x.\int xdx - \int(\frac{d(\cos^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}- \int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
Now, we need to integrate $\int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\cos^{-1}x+\cos^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C$
Put this value in our equation
$\int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}-\left ( \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C \right )\\ \\ \int x\cos^{-1} xdx =\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$

Therefore, the answer is $\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$

### Question:10 Integrate the functions $( \sin ^{-1}x ) ^ 2$

Given function is
$f(x)=( \sin ^{-1}x ) ^ 2$
we will use integration by parts method
$\int (\sin^{-1}x)^2= (\sin^{-1}x)^2.\int 1dx-\int \left ( \frac{d( (\sin^{-1}x)^2)}{dx} .\int 1dx\right )dx\\ \\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x-\int \left ( \sin^{-1}.\frac{2x}{\sqrt{1-x^2}} \right )dx\\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \left ( \frac{d(\sin^{-1}x)}{dx}. \int \frac{-2x}{\sqrt{1-x^2}}dx\right ) \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.2\sqrt{1-x^2}- \int \frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C$
Therefore, answer is $(\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C$

### Question:11 Integrate the functions $\frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}$

Consider $\int \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}dx =I$

So, we have then: $I = \frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}. \cos^{-1}x dx$

After taking $\cos ^{-1}x$ as a first function and $\left ( \frac{-2x}{\sqrt{1-x^2}} \right )$ as second function and integrating by parts, we get

$I =-\frac{1}{2}\left [ \cos^{-1}x\int\frac{-2x}{\sqrt{1-x^2}}dx - \int\left \{ \left ( \frac{d}{dx}\cos^{-1}x \right )\int \frac{-2x}{\sqrt{1-x^2}}dx \right \}dx \right ]$ $=-\frac{1}{2}\left [ \cos^{-1}x.2{\sqrt{1-x^2}} + \int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-\int2dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-2x \right ]+C$

Or,

### Question:12 Integrate the functions $x \sec ^2 x$

Consider $x \sec ^2 x$

So, we have then: $I =\int x\sec^2 x dx$

After taking $x$ as a first function and $\sec^2x$ as second function and integrating by parts, we get

$I =x\int \sec^2 x dx -\int \left \{ \left ( \frac{d}{dx}x \right )\int \sec^2 x dx \right \}dx$

$= x\tan x -\int1.\tan x dx$

$= x\tan x +\log|\cos x | +C$

### Question:13 Integrate the functions $\tan ^{-1} x$

Consider $\tan ^{-1} x$

So, we have then: $I =\int 1.\tan^{-1}x dx$

After taking $\tan^{-1}x$ as a first function and $1$ as second function and integrating by parts, we get

$I = \tan^{-1}x \int 1dx -\int \left \{ \left ( \frac{d}{dx}\tan^{-1}x \right )\int1.dx \right \}dx$

$= \tan^{-1}x.x -\int \frac{1}{1+x^2}.xdx$

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

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

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

### Question:14 Integrate the functions $x ( \log x )^ 2$

Consider $x ( \log x )^ 2$

So, we have then: $I = \int x(\log x)^2 dx$

After taking $(\log x )^2$ as a first function and $x$ as second function and integrating by parts, we get

$I = (\log x )^2 \int xdx -\int \left \{ \left ( \frac{d}{dx} (\log x)^2 \right )\int x.dx \right \}dx$

$= (\log x)^2 .\frac{x^2}{2} - \int \frac{2\log x }{x}.\frac{x^2}{2} dx$

$= (\log x)^2 .\frac{x^2}{2} - \int x\log x dx$

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

### Question:15 Integrate the functions $( x^2 + 1 ) \log x$

Consider $( x^2 + 1 ) \log x$

So, we have then: $I = \int (x^2+1) \log x dx = \int x^2 \log x dx +\int \log x dx$

Let us take $I = I_{1} +I_{2}$ ....................(1)

Where, $I_{1} = \int x^2\log x dx$ and $I_{2} = \int \log x dx$

So, $I_{1} = \int x^2\log x dx$

After taking $\log x$ as a first function and $x^2$ as second function and integrating by parts, we get

$I = \log x \int x^2dx -\int \left \{ \left ( \frac{d}{dx} \log x \right )\int x^2.dx \right \}dx$

$= \log x .\frac{x^3}{3} - \int \frac{1}{x}.\frac{x^3}{3} dx$

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

$I_{2} = \int \log x dx$

After taking $\log x$ as a first function and $1$ as second function and integrating by parts, we get

$I_{2} = \log x \int 1.dx - \int \left \{ \left ( \frac{d}{dx}\log x \right ) \int 1.dx \right \}dx$

$= \log x .x -\int \frac{1}{x}. xdx$

$= x\log x -\int 1 dx$

$= x\log x -x +C_{2}$ ................(3)

Now, using the two equations (2) and (3) in (1) we get,

$I = \frac{x^3}{3}\log x -\frac{x^3}{9} +C_{1} +x\log x - x +C_{2}$

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

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

### Question:16 Integrate the functions $e ^ x ( \sin x + \cos x )$

Let suppose
$I =$ $e ^ x ( \sin x + \cos x )$
$f(x) = \sin x \Rightarrow f'(x) = \cos x$
we know that,
$I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C$
Thus, the solution of the given integral is given by

$\therefore I = e^x\sin x +C$

### Question:17 Integrate the functions $\frac{x e ^x }{( 1+ x )^2}$

$\frac{x e ^x }{( 1+ x )^2}$
Let suppose
$I = \int \frac{e^x(x)}{(1+x)^2}dx$
by rearranging the equation, we get
$\Rightarrow \int e^x[\frac{1}{1+x}-\frac{1}{(1+x)^2}]dx$
let
$f(x)=\frac{1}{1+x} \Rightarrow f'(x)= -\frac{1}{(1+x)^{{2}}}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
therefore the solution of the given integral is

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

### Question:18 Integrate the functions $e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$

Let
$I =e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$
substitute $1 =\sin ^2\frac{x}{2}+\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$

$\\\Rightarrow e^x(\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}})\\ =e^x(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2})\\$
let
$f(x) =\tan\frac{x}{2} \Rightarrow f'(x)=\frac{1}{2}\sec^2\frac{x}{2}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
Therefore the solution of the given integral is

$I = e^x\tan\frac{x}{2} +C$

### Question:19 Integrate the functions $e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$

$e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$
It is known that
$\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

let
$f(x)=\frac{1}{x}\Rightarrow f'(x)=-\frac{1}{x^2}$
Therefore the required solution of the given above integral is
$I = e^x.\frac{1}{x}+C$

### Question:20 Integrate the functions $\frac{( x-3)e ^x }{( x-1)^3}$

$\frac{( x-3)e ^x }{( x-1)^3}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

So, By adjusting the given equation, we get
$\int\frac{( x-3)e ^x }{( x-1)^3} =\int e^x(\frac{x-1-2}{(x-1)^3}) =\int e^x({\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3})}dx$

to let
$f(x)=\frac{1}{(x-1)^2}\Rightarrow f'(x)=-\frac{2}{(x-1)^3}$
Therefore the required solution of the given $I=\frac{e^x}{(x-1)^2}+C$ integral is

### Question:21 Integrate the functions $e ^{ 2x } \sin x$

Let
$I =e ^{ 2x } \sin x$
By using integrating by parts, we get

$\\=\sin x\int e ^{ 2x }dx-\int(\frac{d}{dx}\sin x.\int e^{2x}dx)\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}\int e^{2x}.\cos x\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x\int e^{2x}dx-\int (\frac{d}{dx}\cos x.\int e^{2x}dx)\ dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x.\frac{e^x}{2}+\frac{1}{2}\int e^{2x}\sin x dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}-\frac{1}{4}I\\ \Rightarrow \frac{5}{4}I =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}\\ I = \frac{e^{2x}}{5}[2\sin x-\cos x]+C$

### Question:22 Integrate the functions $\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )$

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

$\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let
$x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$

$\\=\int\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

$\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]+C\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]+C\\ =2x\tan^{-1}x+2\log (1+x^2)^{-1/2}\\ =2x\tan^{-1}x-\log(1+x^2)+C$

### Question:23 Choose the correct answer

$\int x ^ 2 e ^{x ^3 } dx \: \: equals$

$A ) \frac{1}{3} e ^{x^3} + C \\\\ B) \frac{1}{3} e ^{x^2} + C \\\\ C ) \frac{1}{2} e ^{x^3} + C \\\\ D ) \frac{1}{2} e ^{x^2} + C$

the integration can be done ass follows

let $x^3 =t\Rightarrow 3x^2dx=dt$
$\Rightarrow I =\frac{1}{3}\int e^tdt =\frac{1}{3}e^t+C=\frac{1}{3}e^x^3+C$

### Question:24 Choose the correct answer

$\int e ^ x \sec ( 1+ \tan x ) dx \: \: \: equals$

$A ) e ^ x \cos x + C \\\\ B) e ^ x \sec x + C \\\\ C ) e ^ x \sin x + C\\\\D ) e ^ x \tan x + C$

we know that,
$I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C$
from above integral
let
$f(x)=\sec x\Rightarrow f'(x)= \sec x.\tan x$
thus, the solution of the above integral is
$I=e^x\sec x+C$

## More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.6

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Also Read| Integrals Class 12 Notes

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