Exercise 11.13 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.13
Choose the correct or most suitable answer from given four alternatives.
Question 1 .
If $\int f(x) d x=\mathrm{g}(\mathrm{x})+\mathrm{c}$, then $\int f(x) g^{\prime}(x) d x$
(a) $\int(f(x))^2 d x$
(b) $\int f(x) g(x) d x$
(c) $\int f^{\prime}(x) g(x) d x$
(d) $\int(g(x))^2 d x$
Solution:
(a) $\int(f(x))^2 d x$
Hint: $\begin{aligned} \int f(x) d x & =g(x)+c \\ \text { Clearly } g^{\prime}(x) & =f(x) \\ \int f(x) g^{\prime}(x) d x & =\int(f(x))^2 d x\end{aligned}$
Question 2.
If $\int \frac{3^{\frac{1}{x}}}{x^2} d x=k\left(3^{\frac{1}{x}}\right)+c$
(a) $\log 3$
(b) $-\log 3$
(c) $-\frac{1}{\log ^3}$
(d) $\frac{1}{\log 3}$
Solution:
(c) $-\frac{1}{\log ^3}$
$
\begin{aligned}
t & =\frac{1}{x} \\
\frac{d t}{d x} & =\frac{-1}{x^2} \\
-d t & =\frac{d x}{x^2} \\
\int 3^t(-d t) & =-\frac{3^t}{\log 3}=-\frac{3^{\frac{1}{x}}}{\log 3}+c \\
k & =-\frac{1}{\log 3}
\end{aligned}
$

Question 3.
If $\int f^{\prime}(x) e^{x^3} d x=(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}^2}$, then $\mathrm{f}(\mathrm{x})$ is
(a) $2 x^3-\frac{x^2}{2}+x+c$
(b) $\frac{x^2}{2}+3 x^2+4 x+c$
(c) $x^3+4 x^2+6 x+c$
(d) $\frac{2 x^3}{3}-x^2+x+c$
Solution:
(d) $\frac{2 x^3}{3}-x^2+x+c$
$
\text { Hint: } \quad \begin{aligned}
f^{\prime}(x) e^{x^2} & =\frac{d}{d x}\left[(x-1) e^{x^2}\right] \\
& =(x-1) e^{x^2}(2 x)+e^{x^2}(1-0) \\
f^{\prime}(x) e^{x^2} & =e^{x^2}\left[2 x^2-2 x+1\right] \\
f^{\prime}(x) & =2 x^2-2 x+1 \\
f(x) & =\int\left(2 x^2-2 x+1\right) d x+c=\frac{2 x^3}{3}-x^2+x+c
\end{aligned}
$

Question 4.
The gradient (slope) of a curve at any point $(x, y)$ is $\frac{x^2-4}{x^2}$. If the curve passes through the point $(2,7)$, then the equation of the curve is
(a) $y=x+\frac{4}{x}+3$
(b) $y=x+\frac{4}{x}+4$
(c) $y=x^2+3 x+4$
(d) $y=x^2-3 x+6$
Solution:
$
\text { Hint: } \begin{aligned}
\frac{d y}{d x} & =\frac{x^2-4}{x^2} \\
d y & =\left(1-\frac{4}{x^2}\right) d x \\
\int d y & =\int\left(1-\frac{4}{x^2}\right) d x \\
y & =x+\frac{4}{x}+c
\end{aligned}
$
This passes through $(2,7)$
$
\begin{aligned}
& 7=2+\frac{4}{2}+c \\
& c=3 \\
& y=x+4 / x+3
\end{aligned}
$

Question 5.
$\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$ is
(a) $\cot \left(\mathrm{xe}^{\mathrm{x}}\right)+\mathrm{c}$
(b) $\sec \left(x e^x\right)+c$
(c) $\tan \left(x e^x\right)+c$
(d) $\cos \left(x e^x\right)+c$
Solution:
(c)
Hint:
$
\begin{aligned}
& t=x e^x \\
& d t=\left(x e^x+e^x\right) d x \\
& =\int \frac{1}{\cos ^2(t)} d t=\int \sec ^2+d t=\tan t+c \\
& =\tan \left(x e^x\right)+c \\
&
\end{aligned}
$

Question 6.
$\int \frac{\sqrt{\tan x}}{\sin 2 x} d x$ is
(a) $\sqrt{\tan x}+c$
(b) $2 \sqrt{\tan x}+c$
(c) $\frac{1}{2} \sqrt{\tan x}+c$
(d) $\frac{1}{4} \sqrt{\tan x}+c$
Solution:
(a) $\sqrt{\tan x}+c$
$
\begin{aligned}
& \text { Hint: }=\int \frac{\sqrt{\frac{\sin x}{\cos x}}}{2 \sin x \cos x} d x=\int \frac{\sqrt{\sin x}}{2 \sin x \cos x \sqrt{\cos x}} d x \\
& =\int \frac{1}{2 \sqrt{\sin x} \cos x \sqrt{\cos x}} \\
& =\frac{1}{2} \int \frac{1}{\sqrt{\sin x} \cdot \cos x \sqrt{\cos x} \times \frac{\sqrt{\cos x}}{\sqrt{\cos x}}} d x \\
& =\frac{1}{2} \int \frac{1}{\sqrt{\tan x} \cdot \cos ^2 x} d x \\
& =\frac{1}{2} \int \frac{\sec ^2 x}{\sqrt{\tan x}} d x=\frac{1}{2} 2 \sqrt{\tan x}+c \\
& =\sqrt{\tan x}+c \\
&
\end{aligned}
$

Question 7.
$\int \sin ^3 x d x$ is
(a) $\frac{-3}{4} \cos x-\frac{\cos 3 x}{12}+c$
(b) $\frac{3}{4} \cos x+\frac{\cos 3 x}{12}+c$
(c) $\frac{-3}{4} \cos x+\frac{\cos 3 x}{12}+c$
(d) $\frac{-3}{4} \sin x-\frac{\sin 3 x}{12}+c$
Solution:
(c) $\frac{-3}{4} \cos x+\frac{\cos 3 x}{12}+c$
Hint: $\sin ^3 x=\frac{1}{4}(3 \sin x-\sin 3 x)$
$
\begin{aligned}
\int \sin ^3 x d x & =\frac{1}{4} \int(3 \sin x-\sin 3 x) d x \\
& =\frac{1}{4}\left[-3 \cos x+\frac{\cos 3 x}{3}\right]+c \\
& =\frac{-3}{4} \cos x+\frac{\cos 3 x}{12}+c
\end{aligned}
$

Question 8 .
$
\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x} e^{3 \log x}} d x
$
(a) $x+c$
(b) $\frac{x^3}{3}+c$
(c) $\frac{3}{x^3}+c$
(d) $\frac{1}{x^2}+c$
Solution:
(b) $\frac{x^3}{3}+c$
$
\text { Hint: } \begin{aligned}
& =\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x=\int \frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^4}-e^{\log x^3}} d x \\
& =\int \frac{x^6-x^5}{x^4-x^3} d x=\int \frac{x^5(x-1)}{x^3(x-1)} d x=x^2 d x=\frac{x^3}{3}+c
\end{aligned}
$
Question 9.
$\int \frac{\sec x}{\sqrt{\cos 2 x}} d x$ is
(a) $\tan ^{-1}(\sin \mathrm{x})+\mathrm{c}$
(b) $2 \sin ^{-1}(\tan x)+c$
(c) $\tan ^{-1}(\cos x)+c$
(d) $\sin ^{-1}(\tan \mathrm{x})+\mathrm{c}$
Solution:
(d) $\sin ^{-1}(\tan \mathrm{x})+\mathrm{c}$

$\begin{aligned}
& \text { Hint: } \int \frac{\sec x}{\sqrt{\cos 2 x}} d x \\
& =\int \frac{\sec x}{\sqrt{\cos ^2 x-\sin ^2 x}} d x \\
& =\int \frac{\sec x}{\sqrt{\cos ^2 x\left(1-\frac{\sin ^2 x}{\cos ^2 x}\right)}} d x=\int \frac{\sec x}{\cos x \sqrt{1-\tan ^2 x}} d x \\
& =\int \frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}} d x \\
& t=\tan x \\
& d t=\sec ^2 x d x \\
& =\int \frac{d t}{\sqrt{1-t^2}} \\
& =\sin ^{-1}(\mathrm{t})+\mathrm{c} \\
& =\sin ^{-1}(\tan \mathrm{x})+\mathrm{c} \\
&
\end{aligned}$

Question 10.
$
\int \tan ^{-1} \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x
$
(a) $x^2+c$
(b) $2 \mathrm{x}^2+\mathrm{c}$
(c) $\frac{x^2}{2}+c$
(d) $-\frac{x^2}{2}+\mathrm{c}$
Solution:
(c) $\frac{x^2}{2}+c$
$
\text { Hint: } \begin{aligned}
\int \tan ^{-1} \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x & =\int \tan ^{-1} \sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}} d x \\
& =\int \tan ^{-1}(\tan x) d x=\int x d x=\frac{x^2}{2}+c
\end{aligned}
$
Question 11.
$\int 2^{3 x+5} d x$ is ............
(a) $\frac{3\left(2^{3 x+5}\right)}{\log 2}+c$
(b) $\frac{2^{3 x+5}}{2 \log (3 x+5)}+c$

(c) $\frac{2^{3 x+5}}{2 \log 3}+c$
(d) $\frac{2^{3 x+5}}{3 \log 2}+c$

Solution:
(d) $\frac{2^{3 x+5}}{3 \log 2}+c$
Hint: $\quad \int 2^{3 x+5} d x=\frac{2^{3 x+5}}{3 \log 2}+c$
Question 12 .
$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$ is
(a) $\frac{1}{2} \sin 2 x+c$
(b) $-\frac{1}{2} \sin 2 x+c$
(c) $\frac{1}{2} \cos 2 x+c$
(d) $-\frac{1}{2} \cos 2 x+c$
Solution:
(b) $-\frac{1}{2} \sin 2 x+c$
Hint: Let $\quad I=\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$
Consider $\sin ^8 x-\cos ^8 x=\left(\sin ^4 x\right)^2-\left(\cos ^4 x\right)^2$
$=\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^4 x-\cos ^4 x\right)$
$=\left[\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^2+2 \cos ^2 x \sin ^2 x-2 \cos ^2 x \sin ^2 x\right]$
$\left[\sin ^2 x-\cos ^2 x\right]\left[\sin ^2 x+\cos ^2 x\right]$
$=\left[\left(\sin ^2 x+\cos ^2 x\right)^2-2 \cos ^2 x \sin ^2 x\right]\left[\sin ^2 x-\cos ^2 x\right]$
$=\left[1-2 \sin ^2 x \cos ^2 x\right]\left[\sin ^2 x-\cos ^2 x\right]$
$\Rightarrow \quad \mathrm{I}=\int\left(\sin ^2 x-\cos ^2 x\right) d x$
$=\int-\cos 2 x d x$
$=\frac{-\sin 2 x}{2}+c$
Question 13.
$\int \frac{e^x\left(x^2 \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^2+1} d x$ is
(a) $e^x \tan ^{-1}(x+1)+c$
(b) $\tan ^{-1}\left(e^x\right)+c$
(c) $e^x \frac{\left(\tan ^{-1} x\right)^2}{2}+c$
(d) $e^x \tan ^{-1} x+c$
Solution:

(d) $e^x \tan ^{-1} x+c$
Hint: $\int \frac{e^x\left(x^2 \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^2+1} d x=\int e^x\left[\tan ^{-1} x+\frac{1}{x^2+1}\right] d x$
This is of the form of $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c$
$
=e^x \tan ^{-1} x+c
$
Question 14 .
$\int \frac{x^2+\cos ^2 x}{x^2+1} \operatorname{cosec}^2 x d x$ is
(a) $\cot x+\sin ^{-1} x+c$
(b) $-\cot x+\tan ^{-1} x+c$
(c) $-\tan x+\cot ^{-1} x+c$
(d) $-\cot x-\tan ^{-1} \mathrm{x}+\mathrm{c}$
Solution:
(d) $-\cot x-\tan ^{-1} \mathrm{x}+\mathrm{c}$
$
\text { Hint: } \begin{aligned}
\int \frac{x^2+\cos ^2 x}{x^2+1} \operatorname{cosec}^2 x d x & \\
& =\int \frac{\left(x^2+1-\sin ^2 x\right) \operatorname{cosec}^2 x}{x^2+1} d x \\
& =\int \frac{\left(x^2+1\right) \operatorname{cosec}^2 x-1}{x^2+1} d x \\
& =\int \operatorname{cosec}^2 x d x-\int \frac{1}{x^2+1} d x=-\cot x-\tan ^{-1} x+c
\end{aligned}
$

Question 15 .

$\int x^2 \cos x d x$ is
(a) $x^2 \sin \mathrm{x}+2 \mathrm{x} \cos \mathrm{x}-2 \sin \mathrm{x}+\mathrm{c}$
(b) $x^2 \sin x-2 x \cos x-2 \sin x+c$
(c) $-\mathrm{x}^2 \sin \mathrm{x}+2 \mathrm{x} \cos \mathrm{x}+2 \sin \mathrm{x}+\mathrm{c}$
(d) $-\mathrm{x}^2 \sin \mathrm{x}-2 \mathrm{x} \cos \mathrm{x}+2 \sin \mathrm{x}+\mathrm{c}$
Solution:
(a)
Hint: $\int x^2 \cos x d x$
By Bernoullis formula $d v=\cos x d x$
$\mathrm{u}=\mathrm{x}^2 \mathrm{v}=\sin \mathrm{x}$
$\mathrm{u}^{\prime}=2 \mathrm{x}_1=-\cos \mathrm{x}$
$u^{\prime \prime}=2 v_2=-\sin x$
$=u v-u^{\prime} v_1+u^{\prime \prime} v_2$
$=\mathrm{x}^2 \sin \mathrm{x}+2 \mathrm{x} \cos \mathrm{x}-2 \sin \mathrm{x}+\mathrm{c}$

Question 16 .
$\int \sqrt{\frac{1-x}{1+x}} d x$ is
(a) $\sqrt{1-x^2}+\sin ^{-1} x+c$
(b) $\sin ^{-1} x-\sqrt{1-x^2}+c$
(c) $\log \left|x+\sqrt{1-x^2}\right|-\sqrt{1-x^2}+c$
(d) $\sqrt{1-x^2}+\log \left|x+\sqrt{1-x^2}\right|+c$
Solution:
(b) $\sin ^{-1} x-\sqrt{1-x^2}+c$
$
\begin{aligned}
& \text { Hint: Consider, } \quad \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}}=\sqrt{\frac{(1-x)^2}{1-x^2}}=\frac{1-x}{\sqrt{1-x^2}} \\
& \int \frac{1-x}{\sqrt{1-x^2}} d x \Rightarrow 1-x=\mathrm{A} \frac{d}{d x}\left(1-x^2\right)+\mathrm{B} \\
& 1-x=\mathrm{A}(0-2 x)+\mathrm{B} \\
&
\end{aligned}
$
Comparing the like coefficients
$
\begin{aligned}
-2 \mathrm{~A} & =-1, \mathrm{~B}=1 \\
\mathrm{~A} & =\frac{1}{2} \\
\int \frac{1-x}{\sqrt{1-x^2}} d x & =\int \frac{\frac{1}{2}(-2 x)}{\sqrt{1-x^2}} d x+\int \frac{1}{\sqrt{1-x^2}} d x \\
& =-\frac{1}{2} \not 2 \sqrt{1-x^2}+\sin ^{-1}(x)+c \\
& =-\sqrt{1-x^2}+\sin ^{-1}(x)+c
\end{aligned}
$
Question 17.
$\int \frac{d x}{e^2-1}$ is ..
(a) $\log \left|\mathrm{e}^{\mathrm{x}}\right|-\log \left|\mathrm{e}^{\mathrm{x}}-1\right|+\mathrm{c}$
(b) $\log \left|\mathrm{e}^{\mathrm{x}}\right|+\log \left|\mathrm{e}^{\mathrm{x}}-1\right|+\mathrm{c}$
(c) $\log \left|\mathrm{e}^{\mathrm{x}}-1\right|-\log \left|\mathrm{e}^{\mathrm{x}}\right|+\mathrm{c}$
(d) $\log \left|\mathrm{e}^{\mathrm{x}}+1\right|-\log \left|\mathrm{e}^{\mathrm{x}}\right|+\mathrm{c}$
Solution:

(c) $\log \left|\mathrm{e}^{\mathrm{x}}-1\right|-\log \left|\mathrm{e}^{\mathrm{x}}\right|+\mathrm{c}$
Hint: $\int \frac{d x}{e^x-1}$
$
\begin{aligned}
& =\int \frac{1}{e^x-1} d x=-\int \frac{-1}{e^x-1} d x=-\int \frac{e^x-1-e^x}{e^x-1} d x \\
& =-\int\left(1-\frac{e^x}{e^x-1}\right) d x \\
& =-\left[x-\log \left|e^x-1\right|\right]+c \\
& =-x+\log \left|e^x-1\right|+c \\
& =\log \left|e^x-1\right|-x+c \\
& =\log \left|e^x-1\right|-\log \left|e^x\right|+c
\end{aligned}
$
Question 18.
$\int e^{-4 x} \cos x d x$ is
(a) $\frac{e^{-4 x}}{17}[4 \cos x-\sin x]+c$
(b) $\frac{e^{-4 x}}{17}[-4 \cos x+\sin x]+c$
(c) $\frac{e^{-4 x}}{17}[4 \cos x+\sin x]+c$
(d) $\frac{e^{-4 x}}{17}[-4 \cos x-\sin x]+c$
Solution:
(b) $\frac{e^{-4 x}}{17}[-4 \cos x+\sin x]+c$
We know that
$
\text { Hint: } \begin{aligned}
\int e^{a x} \cos b x d x & =\frac{e^{a x}}{a^2+b^2}[a \cos b x+b \sin b x]+c \\
\int e^{-4 x} \cos x d x & =\frac{e^{-4 x}}{16+1}[-4 \cos x+\sin x]+c \\
& =\frac{e^{-4 x}}{17}[-4 \cos x+\sin x]+c
\end{aligned}
$

Question 19.
$
\int \frac{\sec ^2 x}{\tan ^2 x-1} d x
$
(a) $2 \log \left|\frac{1-\tan x}{1+\tan x}\right|+c$
(b) $\log \left|\frac{1+\tan x}{1-\tan x}\right|+c$
(c) $\frac{1}{2} \log \left|\frac{\tan x+1}{\tan x-1}\right|$
(d) $\frac{1}{2} \log \left|\frac{\tan x-1}{\tan x+1}\right|+c$
Solution:

(d) $\frac{1}{2} \log \left|\frac{\tan x-1}{\tan x+1}\right|+c$
HInt: $\int \frac{\sec ^2 x}{\tan ^2 x-1} d x$
Put
$
\begin{aligned}
t & =\tan \mathrm{x} \\
d t & =\sec ^2 x d x \\
& =\int \frac{d t}{t^2-1} \\
& =\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+c \\
& =\frac{1}{2} \log \left|\frac{\tan x-1}{\tan x+1}\right|+c
\end{aligned}
$
Question 20.
$\int e^{-7 x} \sin 5 x d x$ is
(a) $\frac{e^{-7 x}}{74}[-7 \sin 5 x-5 \cos 5 x]+c$
(b) $\frac{e^{-7 x}}{74}[7 \sin 5 x+5 \cos 5 x]+c$
(c) $\frac{e^{-7 x}}{74}[7 \sin 5 x-5 \cos 5 x]+c$
(d) $\frac{e^{-7 x}}{74}[-7 \sin 5 x+5 \cos 5 x]+c$
Solution:
(a) $\frac{e^{-7 x}}{74}[-7 \sin 5 x-5 \cos 5 x]+c$
We know that
$
\text { Hint: } \begin{aligned}
\int e^{a x} \sin b x d x & =\frac{e^{a x}}{a^2+b^2}[a \sin b x-b \cos b x]+c \\
\int e^{-7 x} \sin 5 x d x & =\frac{e^{-7 x}}{49+25}[-7 \sin 5 x-5 \cos 5 x]+c \\
{\left[\begin{array}{c}
a=-7 \\
b=5
\end{array}\right] } & =\frac{e^{-7 x}}{74}[-7 \sin 5 x-5 \cos 5 x]+c
\end{aligned}
$

Question 21.
$\int x^2 e^{\frac{x}{2}} d x$ is
(a) $x^2 e^{\frac{x}{2}}-4 x e^{\frac{x}{2}}-8 e^{\frac{x}{2}}+c$
(b) $2 x^2 e^{\frac{x}{2}}-8 x e^{\frac{x}{2}}-16 e^{\frac{x}{2}}+c$
(c) $2 x^2 e^{\frac{x}{2}}-8 x e^{\frac{x}{2}}+16 e^{\frac{x}{2}}+c$
(d) $x^2 \frac{e^{\frac{x}{2}}}{2}-\frac{x e^{\frac{x}{2}}}{4}+\frac{e^{\frac{x}{2}}}{8}+c$

Solution:
(c)
Hint:
$
I=\int x^2 e^{x / 2} d x
$
$
\begin{aligned}
t & =\frac{x}{2} \Rightarrow x=2 t \\
d t & =\frac{1}{2} d x \\
2 d t & =d x \\
\mathrm{I} & =\int(2 t)^2 e^t(2 d t) \\
& =\int 4 t^2 e^t(2 d t) \\
& =8 \int t^2 e^t d t
\end{aligned}
$
By Bernoullis formula,
$
\begin{aligned}
& =8 e^t\left[t^2-2 t+2\right]+c \\
& =8 e^{x / 2}\left[\frac{x^2}{4}-\frac{2 x}{2}+2\right]+c \\
& =2 x^2 e^{x / 2}-8 x e^{x / 2}+16 e^{x / 2}+c
\end{aligned}
$
Question 22.
$
\int \frac{x+2}{\sqrt{x^2-1}} d x
$
(a) $\sqrt{x^2-1}-2 \log \left|x+\sqrt{x^2-1}\right|+c$
(b) $\sin ^{-1} x-2 \log \left|x+\sqrt{x^2-1}\right|+c$
(c) $2 \log \left|x+\sqrt{x^2-1}\right|-\sin ^{-1} x+c$
(d) $\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1}\right|+c$
Solution:

(d) $\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1}\right|+c$
$
\begin{aligned}
& \int \frac{x+2}{\sqrt{x^2-1}} d x \\
& x+2=\mathrm{A} \frac{d}{d x}\left(x^2-1\right)+\mathrm{B} \\
& x+2=\mathrm{A}(2 x)+\mathrm{B} \\
&
\end{aligned}
$
Comparing the like coefficients of the terms.
$
\begin{aligned}
2 \mathrm{~A} & =1 \quad \mathrm{~B}=2 \\
\mathrm{~A} & =\frac{1}{2} \\
\int \frac{x+2}{\sqrt{x^2-1}} d x & =\int \frac{\frac{1}{2}(2 x)}{\sqrt{x^2-1}}+\int \frac{2}{\sqrt{x^2-1}} d x \\
& =\frac{1}{2} 2 \sqrt{x^2-1}+2\left[\log \left|x+\sqrt{x^2-1}\right|\right]+c \\
& =\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1}\right|+c
\end{aligned}
$
Question 23.
$
\int \frac{1}{x \sqrt{(\log x)^2-5}} d x
$
(a) $\log \left|x+\sqrt{x^2-5}\right|+c$
(b) $\log |\log x+\sqrt{\log x-5}|+c$
(c) $\log \left|\log x+\sqrt{(\log x)^2-5}\right|+c$
(d) $\log \left|\log x-\sqrt{(\log x)^2-5}\right|+c$
Solution:
(c) $\log \left|\log x+\sqrt{(\log x)^2-5}\right|+c$

$\begin{aligned}
& I=\int \frac{1}{x \sqrt{(\log x)^2-5}} d x \\
& \text { Put } \\
& t=\log x \\
& e^t=x \\
& e^t d t=d x \\
& I=\int \frac{1}{\sqrt{t^2-5}} e^t d t \\
& =\int \frac{1}{e^t \sqrt{t^2-(\sqrt{5})^2}} e^t d t \\
& =\log \left|t+\sqrt{t^2-(\sqrt{5})^2}\right|+c \\
& =\log \left|\log x+\sqrt{(\log x)^2-5}\right|+c \\
&
\end{aligned}$

Question 24.
$
\int \sin \sqrt{x} d x=
$
(a) $2(-\sqrt{x} \cos \sqrt{x}+\sin \sqrt{x})+c$
(b) $2(-\sqrt{x} \cos \sqrt{x}-\sin \sqrt{x})+c$
(c) $2(-\sqrt{x} \sin \sqrt{x}-\cos \sqrt{x})+c$
(d) $2(-\sqrt{x} \sin \sqrt{x}+\cos \sqrt{x})+c$
Solution:
(a) $2(-\sqrt{x} \cos \sqrt{x}+\sin \sqrt{x})+c$
$
\text { Hint: } \begin{aligned}
\mathrm{I} & =\int \sin \sqrt{x} d x \\
t & =\sqrt{x} \\
d t & =\frac{1}{2 \sqrt{x}} d x \\
2 \sqrt{x} d t & =d x \\
2 t d t & =d x \\
\mathrm{I} & =\int \sin t(2 t d t) \\
& =2 \int t \sin t d t \\
\int u d v & =u v-\int v d u \\
& =2\left[t(-\cos t)-\int(-\cos t) d t\right]+c \\
& =2[-t \cos t+\sin t]+c \\
& =2[-\sqrt{x} \cos \sqrt{x}+\sin \sqrt{x}]+c
\end{aligned}
$

Question 25 .
$\int e^{\sqrt{x}} d x$ is
(a) $2 \sqrt{x}\left(1-e^{\sqrt{x}}\right)+c$
(b) $2 \sqrt{x}\left(e^{\sqrt{x}}-1\right)+c$
(c) $2 e^{\sqrt{x}}(1-\sqrt{x})+c$
(d) $2 e^{\sqrt{x}}(\sqrt{x}-1)+c$
Solution:
(d)
Hint: Let $\mathrm{I}=\int e^{\sqrt{x}} d x$
$
\mathrm{t}=\sqrt{x}
$
$
\begin{aligned}
d t & =\frac{1}{2 \sqrt{x}} d x \\
2 \sqrt{x} & =d x \\
2 t d t & =d x \\
\mathrm{I} & =\int e^t 2 t d t \\
& =2 \int t e^t d t
\end{aligned}
$
Using Bernoulli's formula
$
\begin{aligned}
& =2 e^t[t-1]+c \\
& =2 e^{\sqrt{x}}[\sqrt{x}-1]+c
\end{aligned}
$