NCERT Solutions for Exercise 7.4 Class 12 Maths Chapter 7 - Integrals

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

Answer:

$\frac{1}{\sqrt { 1+ 4 x^2 }}$
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 $\int\frac{1}{\sqrt{x^2+a^2}}dt = \log\left | x+\sqrt{x^2+a^2} \right |$

Question3: Integrate the function $\frac{1}{\sqrt { ( 2- x)^2+ 1 }}$

Answer:

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

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

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

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

Question4: Integrate the function $\frac{1}{\sqrt {9 - 25 x^2 }}$

Answer:

$\frac{1}{\sqrt {9 - 25 x^2 }}$
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 function $\frac{3x }{1+ 2 x ^ 4 }$

Answer:

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

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

Answer:

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

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

Answer:
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 $\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}$

Answer:

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

Answer:

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

Answer:

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

Answer:

$\frac{1}{9 x ^2 + 6x + 5 }$
this denominator can be written as
$9x^2+6x+5=9[x^2+\frac{2}{3}x+\frac{5}{9}]\\=9[(x+\frac{1}{3})^2+(\frac{2}{3})^2]$ Now,
$\frac{1}{9}\int \frac{1}{(x+\frac{1}{3})^2+(\frac{2}{3})^2}dx =\frac{1}{9} [\frac{3}{2}\tan^{-1}(\frac{(x+1/3)}{2/3})] +C\\=\frac{1}{6} \tan^{-1}(\frac{3x+1}{2})] +C$
......................................by using the form $(\int \frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a}))$

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

Answer:

The denominator can also be 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 function $\frac{1}{\sqrt { ( x-1)( x-2 )}}$

Answer:

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

Answer:

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 $x-3/2 = t \Rightarrow dx =dt$

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

Question15: Integrate the function $\frac{1}{\sqrt {(x-a)( x-b )}}$

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

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

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

Answer:

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 $\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals$

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

Answer:

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)