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

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

Answer:

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

$\int \frac{1}{\sqrt{1+4x^2}}=\frac{1}{2}\int \frac{dt}{1+t^2}$
$$\begin{gathered} =\frac{1}{2}[\log |t+\sqrt{1+t^2}|]+C \\ =\frac{1}{2}\log |2x+\sqrt{4x^2+1}|+C \end{gathered}$$ .................using formula $\int \frac{1}{\sqrt{x^2+a^2}}dt=\log |x+\sqrt{x^2+a^2}|$

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

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

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

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

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

Answer:

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

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

The above result is obtained using the identity

$$\begin{gathered} \int \frac{1}{\sqrt{a^2-x^2}}dt \\ =\frac{1}{a}si{n}^{-1}\frac{x}{a} \end{gathered}$$

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

Answer:

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

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

The integration can be done as follows

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

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

Answer:

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

let $x^3=t$
then $3x^{2}dx=dt$

using the special identities we can simplify the integral as follows

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

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

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 $x^2-1=t\Rightarrow 2xdx=dt$

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

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 $x^3=t\Rightarrow 3x^{2}dx=dt$

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

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

Answer:

The integral can be evaluated as follows

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

$\Rightarrow \int \frac{{\sec }^{2}x}{\sqrt{{\tan }^{2}x+4}}dx=\int \frac{dt}{\sqrt{t^2+2^2}}$
$$\begin{gathered} =\log |t+\sqrt{t^2+4}|+C \\ =\log |\tan x+\sqrt{ta{n}^{2}x+4}|+C \end{gathered}$$

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

Answer:

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

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

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

Answer:

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

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

Answer:

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 $\int \frac{1}{\sqrt{a^2-x^2}}={\sin }^{-1}(\frac{x}{a})$
$$\begin{gathered} =si{n}^{-1}(\frac{t}{4})+C \\ ={\sin }^{-1}(\frac{x+3}{4})+C \end{gathered}$$

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

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 $\int \frac{1}{\sqrt{x^2-a^2}}=\log |x+\sqrt{x^2+a^2}|$
$$\begin{gathered} =\log |t+\sqrt{t^2-(1/2{)}^2}|+C \\ =\log |(x-\frac{3}{2})+\sqrt{x^2-3x+2}|+C \end{gathered}$$

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

Answer:

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

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

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

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

Answer:

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

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

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

Answer:

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

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

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

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

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

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

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

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

Answer:

let
$$\begin{gathered} 5x+2=A\frac{d}{dx}(1+2x+3x^2)+B \\ 5x+2=A(2+6x)+B=2A+B+6Ax \end{gathered}$$
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
$I_1=\frac{5}{6}\int \frac{dt}{t}=\frac{5}{6}\log t=\frac{5}{6}\log (3x^2+2x+1)+c1$
$I_2=\int \frac{dx}{3x^2+2x+1}=\frac{1}{3}\int \frac{dx}{(x+1/3{)}^2+(\sqrt{2}/3{)}^2}$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}(\frac{3x+1}{\sqrt{2}})+c2$

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

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

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

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

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

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

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

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

$$\begin{gathered} \therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx=-\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}+4\int \frac{dx}{\sqrt{4x-x^2}} \\ I=\frac{-1}{2}{I}_1+4I_2 \end{gathered}$$ ....................(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 $\frac{x+2}{\sqrt{x^2+2x+3}}$

Answer:

$\frac{x+2}{\sqrt{x^2+2x+3}}$
$\int \frac{x+2}{\sqrt{x^2+2x+3}}dx=\frac{1}{2}\int \frac{2(x+2)}{\sqrt{x^2+2x+3}}dx$
$$\begin{gathered} =\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\frac{1}{2}\int \frac{2}{\sqrt{x^2+2x+3}}dx \\ =\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\int \frac{1}{\sqrt{x^2+2x+3}}dx \\ I=\frac{1}{2}{I}_1+I_2 \end{gathered}$$ ...........(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 $I_2$

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

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

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

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

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

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

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

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

$I=I_1+I_2$

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

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

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)

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

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

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

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

Question:24 Choose the correct answer

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

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

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, $\int \frac{dx}{(x+1{)}^2+1}=ta{n}^{-1}(x+1)+C$

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

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

Answer:

The following integration can be done as

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

The correct option is (B)