KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals Exercise 19.8

Exercise 19.8

Question 1
$\int \frac{\cos x}{(1+\sin x)^{2}} d x$
Sol :
Let z=1+sinx then dz=cosxdx

Now , $\int \frac{\cos x}{(1+\sin^2 x)^{2}} d x$

$=\int \frac{d z}{z^{2}}$

$=\int z^{-2} d z$

$=\frac{z^{-2+1}}{-2+1}+c$

$=\frac{z^{-1}}{-1}+c$

$=-\frac{1}{z}+c$

$=-\frac{1}{(1+\sin x)}+c$


Question 2
$\int \frac{\sin x d x}{1+\cos x}$
Sol :
Let z=+cosx then dz=-sindx
⇒-dz=sinxdx

Now , $\int \frac{\sin x}{1+\cos x} d x$

$=\int \frac{-d z}{z}$

$=-\int \frac{1}{z} d z$

=-logz+c

=-log(1+cosx)+c


Question 3
$\int \frac{\sin x}{(1+\cos x)^{2}} d x$
Sol :
Let z=(1+cosx) then dz=-sinxdx
⇒-dz=sinxdx

Now ,$\int \frac{\sin x}{(1+\cos x)^{2}} d x$

$=-\int \frac{d z}{z^{2}}=\int z^{-2} d z$

$=\frac{-z^{-2+1}}{-2+1}+c$

$=-\frac{z^{-1}}{-1}+c=z^{-1}+c$

$=\frac{1}{z}+c=\frac{1}{(1+\cos x)}+c$


Question 4
$\int \frac{1+\cos x}{\sqrt[3]{x+\sin x}} d x$
Sol :
Let z=x+sinx then
dz=(+cosx)dx

Now , $\int \frac{(1+\cos x) d x}{\sqrt[3]{x+\sin x}}$

$=\int \frac{d z}{\sqrt[3]{z}}=\int \frac{1}{z^{\frac{1}{3}}} d z$

$=\int z^{-\frac{1}{3}} d z$

$=\frac{z^{-\frac{1}{3}+1}}{\frac{-1}{3}+1}+c$

$=\frac{z^{\frac{-1+3}{3}}}{\frac{-1+3}{3}}+c$

$=\frac{z^{\frac{2}{3}}}{\frac{2}{3}}+c$

$=\frac{3}{2} z^{\frac{2}{3}}+c$

$=\frac{3}{2}(x+\sin x)^{\frac{2}{3}}+c$



Question 5
$\int \frac{\sec ^{2} x d x}{\sqrt{1+\tan x}}$
Sol :
Let z=1+tanx then $dz=\sec ^{2} x d x$

Now , $\int \frac{\sec ^{2} x d x}{\sqrt{1+\tan x}}=\int \frac{d z}{\sqrt{z}}$

$=\int z^{-\frac{1}{2}} d z=\frac{z^{-\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=\frac{z^{\frac{1}{2}}}{\frac{1}{2}}+c=2 z^{\frac{1}{2}}+c$

$=2 \sqrt{z}+c$

$=2 \sqrt{1+\tan x}+c$


Question 6
$\int \frac{\sin x}{a+b \cos x} d x$
Sol :
Let z=a+bcosx then dz=b(-sinx)dx
dz=-bsinxdx
$-\frac{d z}{b}=\sin x d x$

Now , $\int \frac{\sin x d x}{a+b \cos x}$

$=\int \frac{-d z}{b \cdot z}=-\frac{1}{b} \int \frac{1}{z} d z$

$=\frac{-1}{b} \log z+c$

$=-\frac{1}{b} \log |a+b \cos x|+c$

Question 7
(i) $\int \frac{\cos x}{1+\sin ^{2} x} d x$
Sol :
Let z=sinx then dz=cosxdx

Now , $\int \frac{\cos x d x}{1+\sin ^{2} x}=\int \frac{d z}{1+z^{2}}$

$=\tan ^{-1} z+c$

$=\tan ^{-1}(\sin x)+c$


(ii) $\int \cos 6 x \cdot \sqrt{1+\sin 6 x} d x$
Sol :
Let z=1+sin6x then dz=6cos6x
⇒$\frac{d z}{6}=\cos 6 x d x$

Now ,$\int \cos 6 x \cdot \sqrt{1+\sin 6 x} d x$

$=\int \sqrt{1+\sin 6 x} \cdot \cos 6 x d x$

$=\int \sqrt{z} \frac{d z}{6}=\frac{1}{6} \int z^{\frac{1}{2}} d z$

$=\frac{1}{6} \cdot \frac{z^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=\frac{1}{6} \cdot \frac{z^{\frac{3}{2}}}{\frac{3}{2}}+c$

$=\frac{1}{6} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=\frac{1}{9} z^{\frac{3}{2}}+c$

$=\frac{1}{9}(1+\sin 6 x)^{\frac{3}{2}}+c$

Question 8
(i) $\int \frac{\sin x+\cos x}{\sin x-\cos x} d x$
Sol :
Let z=sinx-cosx then dz=(cosx+sinx)dx

Now , $\int \frac{\sin x+\cos x}{\sin x-\cos x} d x$

$=\int \frac{\cos x+\sin x}{\sin x-\cos x} d x$

$=\int \frac{d z}{2}$

=logz+c

=log|sinx-cosx|+c


(ii) $\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x$
Sol :
Let z=6cosx+4sinx then
dz=6(-sinx)+4cosxdx
dz=(4cosx-6sinx)dx
dz=2(2cosx-3sinx)dx

$\frac{d z}{2}=(2 \cos x-3 \sin x) d x$

Now , $\int \frac{d z}{2 \cdot z}$

$=\frac{1}{2} \int z d z$

$=\frac{1}{2} \int z d z$

$=\frac{1}{2} \log z+c$

$=\frac{1}{2} \log |6 \cos x+4 \sin x|+c$

Question 9
(i) $\int \frac{\sec ^{2} x d x}{\sqrt{1-\tan ^{2} x}}$
Sol :
Let z=tanx then $dz=\sec ^{2} x d x$

Now , $\int \frac{\sec ^{2} x d x}{\sqrt{1-\tan ^{2} x}}$

$=\int \frac{d z}{\sqrt{1-z^{2}}}$

$=\sin ^{-1} z+c$

$=\sin ^{-1}(\tan x)+c$

(ii) $\int \frac{d x}{\cos ^{2} x(1-\tan x)^{2}}$
Sol :
Let z=tanx then $dz=\sec ^{2} x d x$

Now , $\int \frac{d x}{\cos ^{2} x(1-\tan x)^{2}}$

$=\int \frac{\sec ^{2} x d x}{(1-\tan x)^{2}}=\int \frac{d z}{(1-z)^{2}}$

$=\int(1-z)^{-2} d z$

$=\frac{(-1)(1-z)^{-2+1}}{-2+1}+c$

$=\frac{-1 \times(1-2)^{-2+1}}{-1}+c$

$=(1-z)^{-1}+c$

$=\frac{1}{1-z}+c$

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



Question 10
(i) $\int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x$
Sol :
$=\int \frac{\cos 2 x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cdot \cos x} d x$

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

Let z=1+sin2x then dz=2cos2xdx

$\frac{d z}{2}=\cos 2 x d x$

Now , $\int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x$

$=\int \frac{\cos 2 x}{1+\sin 2 x} d x=\int \frac{d z}{2 \cdot z}$

$=\frac{1}{2} \int \frac{1}{z} d z$

$=\frac{1}{2} \log z+c$

$=\frac{1}{2} \log |1+\sin 2 x|+c$


(ii) $\int \frac{\cos x-\sin x}{1+\sin 2 x} d x$
Sol :
$=\int \frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cdot \cos x} d x$

$=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$

Let z=sinx+cosx then dz=(cosx-sinx)dx

Now, $\int \frac{(\cos x-\sin x) d x}{1+\sin 2 x}$

$=\int \frac{(\cos x-\sin x) d x}{(\sin x+\cos x)^{2}}$

$=\int \frac{d z}{z^{2}}=\int z^{-2} d z$

$=\frac{z^{-2+1}}{-2+1}+c=\frac{z^{-1}}{-1}+c$

$=-\frac{1}{z}+c$

$=\frac{-1}{\sin x+\cos x}+c$

$=-\frac{1}{\cos x+\sin x}+c$


Question 11
(i) $\int(\cot x+x) \cdot \cot ^{2} x d x$
Sol :
Let z=cotx+x then $d z=-\operatorname{cosec}^{2} x+1$
$=\left(1-\operatorname{cosec}^{2} x\right) d x$

$d z=-\cot ^{2} x d x$

$-dz=\cot ^{2} x d x$

Now , $\int(\cot x+x) \cdot \cot ^{2} x d x$

$=\int z \cdot(-d z)$

$=-\int z d z=\frac{-z^{2}}{2}+c$

$=-\frac{(x+\cot x)^{2}}{2}+c$


(ii) $\int \sqrt{\sin 2} x \cdot \cos 2 x d x$
Sol :
Let z=sin2x then dz=2cos2xdx

$\frac{d 2}{2}=\cos 2 x d x$

Now , $\int \sqrt{\sin 2 x} \cdot \cos 2 x d x$

$=\int \sqrt{2} \cdot \frac{d z}{2}=\frac{1}{2} \int \sqrt{z} d z$

$=\frac{1}{2} \int z^{\frac{1}{2}} d z$

$=\frac{1}{2} \cdot \frac{z^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=\frac{1}{2} \frac{2}{3} z^{3/8}$

$=\frac{1}{3} z^{\frac{3}{2}}+c$

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

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



Question 12
$\int(a+b \cos x)^{n} \sin x d x$
Sol :
Let z=(a+bcosx) then dz=b(-sinx)dx =-bsinxdx
$-\frac{d z}{b}=\sin x d x$

Now , $\int(a+b \cos x)^{2} \cdot \sin x d x$
$=\int z^{n}\left(-\frac{d z}{b}\right)=-\frac{1}{b} \int z^{n} d z$

$=-\frac{1}{b} \frac{z^{n+1}}{n+1}+c$

$=-\frac{1}{b} \cdot \frac{(a+b \cos x)^{n+1}}{n+1}+c$



Question 13
(i) $\int(2 x+3) \cdot \sqrt{x^{2}+3 x} d x$
Sol :
Let $z=x^{2}+3 x$ then dz=(2x+3)dx

Now , $\int(2 x+3) \cdot \sqrt{x^{2}+3 x} d x$

$=\int \sqrt{x^{2}+3 x} \cdot(2 x+3) d x$

$=\int \sqrt{z} \cdot d z=\int z^{\frac{1}{2}} d z$

$=\frac{z^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=\frac{2}{3} z^{\frac{3}{2}}+c$

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

(ii) $\int(4 x+2) \sqrt{x^{2}+x+1} d x$
Sol :
$=\int 2(2 x+1) \sqrt{x^{2}+x+1} d x$

$=2 \int \sqrt{x^{2}+x+1} \cdot(2 x+1) d x$

Let $z=x^{2}+x+1$ then dz=(2x+1)dx

Now , $\int(4 x+2) \sqrt{x^{2}+x+1} d x$

$=2 \int \sqrt{x^{2}+x+1} \cdot(2 x+1) d x$

$=2 \int \sqrt{z} \cdot d z$

$=2 \int z^{\frac{1}{2}} d z$

$=2 \cdot \frac{z^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=2 .\frac{2}{3} z^{\frac{3}{2}}+c$

$=\frac{4}{3} z^{\frac{3}{2}}+c$

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


Question 14
$\int(4 x+3) \sqrt{4 x^{2}+6 x+1} d x$
Sol :
Let $z=4 x^{2}+6 x+1$ then
dz=(8x+6)dx
dz=2(4x+3)dx
dz=2(4x+3)dx

$\frac{d z}{2}=(4 x+3) d x$

Now , $\int(4 x+3) \sqrt{4 x^{2}+6 x+1} d x$

$=\int \sqrt{4 x^{2}+6 x+1} \cdot(4 x+3) d x$

$=\int \sqrt{z} \cdot \frac{d z}{2}=\frac{1}{2} \int z^{\frac{1}{2}} d z$

$=\frac{1}{2} \cdot \frac{z^{\frac{1}{2}+1} }{\frac{1}{2}+1}+c$

$=\frac{1}{2}\times \frac{2}{3} z^{\frac{3}{2}}+c$

$=\frac{1}{3} z^{\frac{3}{2}}+c$

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

Question 15
(i) $\int x \cdot \sqrt{1+x^{2}} d x$
Sol :
Let $z=1+x^{2}$ then $d z=2 x d x$

⇒$\frac{d z}{2}=x d x$

Now , $\int x \sqrt{1+x^{2}} d x$

$=\int \sqrt{1+x^{2}} \cdot x d x=\int \sqrt{z} \cdot \frac{d z}{2}$

$=\frac{1}{2} \int z^{\frac{1}{2}} dz$

$=\frac{1}{2} \cdot \frac{z^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=\frac{1}{2} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=\frac{1}{3} z^{\frac{3}{2}}+c$

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


(ii) $\int x \cdot \sqrt{1+2 x^{2}} d x$
Sol :
Let $z=1+2 x^{2}$ then
dz=2.2xdx
dz=4xdx

$\frac{d z}{4}=x d x$

Now , $\int x \cdot \sqrt{1+2 x^{2}} d x$

$=\int \sqrt{1+2 x^{2}} \cdot x d x=\int \sqrt{z} \cdot \frac{d z}{4}$

$=\frac{1}{4} \int z^{\frac{1}{2}} d z$

$=\frac{1}{4} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}+c$

$=\frac{1}{6} z^{\frac{3}{2}}+c$

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


Question 16
$\int \frac{3 x-1}{\sqrt{3 x^{2}-2 x+7}} d x$
Sol :
Let $z=3 x^{2}-2 x+7$ then
dz=(6x-2)dx
dz=2(3x-1)dx
dz=2(3x+1)dx
$\frac{d z}{2}=(3 x-1) d x$

Now , $\int \frac{3 x-1}{\sqrt{3 x^{2}-2 x+7}} d x$

$\int \frac{1}{\sqrt{z}} \cdot \frac{d z}{2}$

$=\frac{1}{2} \int z^{-\frac{1}{2}} d z$

$=\frac{1}{2} \cdot \frac{z^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}+c$

$=\frac{1}{2} \cdot 2 \cdot z^{\frac{1}{2}}+c$

$=\sqrt{z}+c$

$=\sqrt{3 x^{2}-2 x+7}+c$


Question 17
$\int(x-1) \sqrt{x^{2}-2 x} d x$
Sol :
Let
$z=x^{2}-2 x$ then
dz=(2x-2)dx
dz=2(x-1)dx

$\frac{d z}{2}=(x-1) d x$

Now , $\int(x-1) \sqrt{x^{2}-2x} d x$

$=\int \sqrt{x^{2}-2 x} \cdot(x-1) d x$

$=\int \sqrt{z} \cdot \frac{d z}{2}=\frac{1}{2} \int z^{\frac{1}{2}} d z$

$=\frac{1}{2} \cdot \frac{2}{3} z^{\frac{3}{2}}+c$

$=\frac{1}{2}(z)^{\frac{3}{2}}+c$

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


Question 18
(i) $\int \frac{2 a x+b}{a x^{2}+b x+c} d x$
Sol :
Let $z=a x^{2}+b x+c$ then $(2 a x+b) d x=d z$

Now , $\int \frac{(2 a x+b) d x}{a x^{2}+b x+c}$

$=\int \frac{d z}{z}$

=log|z|+c

$=\log \left|a x^{2}+b x+c\right|+k$


(ii) $\int \frac{2 x}{1+x^{2}} d x$
Sol :
Let $z=1+x^{2}$ then dz=2xdx

Now ,  $\int \frac{2 x d x}{1+x^{2}}=\int \frac{d z}{z}$

=logz+c

$=\log \left|1+x^{2}\right|+c$


Question 19
$\int \frac{2 a x+b}{\sqrt{a x^{2}+b x+c}} d x$
Sol :
Let $z=a x^{2}+b x+c$ then dz=(2ax+b)dx

Now , $\int \frac{(2 a x+b) d x}{\sqrt{a x^{2}+b x+c}}=\int \frac{d z}{\sqrt{z}}$

$=\int z^{-\frac{1}{2}} d z$

$=\frac{z^{-1+1}}{\frac{-1}{2}+1}+c$

$=2 z^{\frac{1}{2}}+c$

$=2 \sqrt{z}+c$

$=2 \sqrt{a x^{2}+b x+c}+k$

Question 20
$\int x \cdot\left(1+3 x^{2}\right)^{\frac{5}{2}} d x$
Sol :
Let $z=1+3 x^{2}$ then dz=6xdx

⇒$\frac{d z}{6}=x d x$

Now , $\int x \cdot\left(1+3 x^{2}\right)^{\frac{5}{2}} d x$

$=\int\left(1+3 x^{2}\right)^{\frac{5}{2}} x d x$

$=\int z^{\frac{5}{2}} \frac{d z}{6}$

$=\frac{1}{6} \int z^{\frac{5}{2}} d z$

$=\frac{1}{6} \cdot \frac{z^{\frac{5}{2}+1}}{\frac{5}{2}+1}+c$

$=\frac{1}{6} .\frac{2}{7} z^{\frac{7}{2}}+c$

$=\frac{1}{21} z^{\frac{7}{2}}+c$

$=\frac{1}{21}\left(1+3 x^{2}\right)^{\frac{7}{2}}+c$


Question 21
(i) $\int \frac{x^{2}}{x^{3}+1} d x$
Sol :
Let $z=x^{3}+1$then $d z=3 x^{2} d x$

Now , $\int \frac{x^{2}}{x^{3}+1} d x$

$=\int \frac{d z}{3 z}=\frac{1}{3} \int \frac{1}{z} d z$

$=\frac{1}{3} \log |z|+c$

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


(ii) $\int \frac{x^{2} d x}{\left(2 x+3 x^{3}\right)^{3}}$

$\int \frac{x^{2} d x}{\left(2+3 x^{3}\right)^{3}}$
Sol :
Let $z=2+3 x^{3}$ then
$d z=3 \cdot 3 x^{2} d x=9 x^{2} d x$

$\frac{d z}{9}=x^{2} d x$

Now , $\int \frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}=\int \frac{d z}{9 \cdot z^{3}}$

$=\frac{1}{9} \int \frac{1}{z^{3}} d z=\frac{1}{9} \int z^{-3} d z$

$=\frac{1}{9} \frac{z^{-3+1}}{-3+1}+c=\frac{1}{9} \cdot \frac{z^{-2}}{-2}+c$

$=\frac{-1}{18} \cdot z^{-2}+c$

$=-\frac{1}{18\times z^{2}}+c$\

$=\frac{-1}{18\left(2+3 x^{3}\right)^{2}}+c$


(iii) $\int \frac{x d x}{\sqrt{a^{2}-x^{2}}}$
Sol :
Let $z=a^{2}-x^{2}$ then

$d z=-2 x d x \Rightarrow-\frac{d z}{2}=x d x$

Now , $\int \frac{x d x}{\sqrt{a^{2}-x^{2}}}=\int \frac{-d z}{2 \sqrt{z}}$

$=\frac{-1}{2} \int \frac{1}{\sqrt{z}} d z=-\frac{1}{2} \int z^{-\frac{1}{2}} d z$

$=-\frac{1}{2} \frac{z^{-\frac{1}{2}+1}}{\frac{1}{2}+1}+c=\frac{-1}{2} \frac{z^{\frac{1}{2}}}{\frac{1}{2}}+c$

$=-z^{\frac{1}{2}}+c$

$=-\sqrt{z}+c$

$=-\sqrt{a^{2}-x^{2}}+c$

Question 23
$\int \frac{x d x}{\sqrt{a^{2}+x^{2}}}$
Sol :
Let $z=a^{2}+x^{2}$ then

$d z=2 x d x \quad \Rightarrow \frac{d z}{2}=x d x$

Now , $\int \frac{x}{\sqrt{a^{2}+x^{2}}} d x=\int \frac{1}{\sqrt{z}} \cdot \frac{d z}{2}$

$=\frac{1}{2} \int \frac{1}{\sqrt{z}} d z$

$=\frac{1}{2} \int z^{-\frac{1}{2}} d z$

$=\frac{1}{2} \frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c$

$=\frac{1}{2} \frac{z^{\frac{1}{2}}}{\frac{1}{2}}$

$=z^{\frac{1}{2}}+c=\sqrt{z}+c$

$=\sqrt{a^{2}+x^{2}}+c$


Question 24
$\int \frac{x}{\left(a x^{2}+b\right)^{p}} d x$
Sol :
Let $z=a x^{2}+b$ then

$d z=a \cdot 2 x d x=2 a x d x$
$\frac{d z}{2 a}=x d x$

Now , $\int \frac{x}{\left(a x^{2}+b\right)^p} d x=\int \frac{1}{z^{p}} \cdot \frac{d z}{2 a}$

$=\frac{1}{2 a} \int z^{-p} d z$

$=\frac{1}{2 a} \cdot \frac{z^{-p+1}}{-p+1}+c$

$=\frac{1}{2 a} \frac{z^{1-p}}{1-p}+c$

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

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



Question 25
(i) $\int x^{3} \cdot \sqrt{x^{2}+5} d x$
Sol :
Let $z=x^{2}+5$ then dz=2xdx

$\Rightarrow \frac{d z}{2}=x d x$

Also , $z=x^{2}+5 \therefore x^{2}=z-5$

Now , $\int x^{2} \sqrt{x^{2}+5} d x=\int x^{2} \cdot x \sqrt{x^{2}+5} d x$

$=\int x^{2} \sqrt{x^{2}+5} x d x$

$=\int(2-5) \cdot \sqrt{z} \frac{d z}{2}=\frac{1}{2} \int(z-5) \sqrt{z} d z$

$=\frac{1}{2} \int\left(z^{\frac{3}{2}}-5z^{\frac{1}{2}}\right) d z$

$=\frac{1}{2}\left[\frac{2}{5} z^{\frac{5}{2}}-5 \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}\right]+c$

$=\frac{1}{2}\left[z^{\frac{3}{2}}\left(\frac{2}{5} z-\frac{10}{3}\right)\right]+c$

$=\frac{1}{2} z^{\frac{3}{2}}\left(\frac{(6z-50)}{15}\right)+c$

$=\frac{1}{2} z^{\frac{3}{2}} \frac{2(3z-25)}{15}+c$

$=\frac{(3z-25) \cdot z^{\frac{3}{2}}+c}{15}$

$=\frac{\left[3\left(x^{2}+5\right)-25\right]}{15} \cdot\left(x^{2}+5\right)^{\frac{3}{2}}+c$

$=\frac{3 x^{2}+15-25}{15} \cdot\left(x^{2}+5\right)^{\frac{3}{2}}+c$

$=\frac{1}{15}\left(3 x^{2}-10\right) \cdot\left(x^{2}+5\right)^{\frac{3}{2}}+c$


(ii) $\int x^{2} \sqrt{a^{3}+x^{3}} d x$
Sol :
Let $z=a^3+x^3$ then
$d z=3 x^{2} d x \Rightarrow \frac{d z}{3}=x^{2} d x$

Now  , $\int x^{2} \cdot \sqrt{a^{2}+x^{3}} d x$

$=\int \sqrt{a^{3}+x^{3}} \cdot x^{2} d x$

$=\int \sqrt{z} \cdot \frac{d z}{3}$

$=\frac{1}{3} \int \sqrt{z} d z$

$=\frac{1}{3} \frac{z^{\frac{1}{2}+1} }{\frac{1}{2}+1}+c$

$=\frac{1}{3} \cdot \frac{2}{3} z^{\frac{3}{2}}+c$

$=\frac{2}{9}\left(a^{3}+x^{3}\right)^{\frac{3}{2}}+c$


Question 26
(i) $\int\left(x^{3}-1\right)^{\frac{1}{3}} \cdot x^{5} d x$
Sol :
Let $z=x^{3}-1$ then

$d z=3 x^{2} d x+\frac{d z}{3}=x^{2} d x$

Also , $z=x^{3}-1 \quad \therefore x^{3}=z+1$

Now , $\int\left(x^{3}-1\right)^{\frac{1}{3}} x^{5} d x$

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

$=\int z^{\frac{1}{2}}(z+1) \frac{d z}{3}$

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

$=\frac{1}{3}\left[\frac{3}{7} \cdot 2^{\frac{7}{3}}+\frac{3}{4} \cdot z^{\frac{4}{3}}\right]+c$

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

$=\frac{1}{3} z^{\frac{4}{3}}\left[\frac{12z+21}{28}\right]+c$

$=\frac{1}{3}\left(x^{3}-1\right)^{\frac{4}{3}} \cdot\left[\frac{12\left(x^{3}-1\right)+21}{28}\right]+c$

$=\frac{1}{3}\left(x^{3}-1\right)^{\frac{4}{3}}\left[\frac{12 x^{3}-12+21}{28}\right]+c$

$=\frac{1}{3}\left(x^{3}-1\right)^{\frac{4}{3}}\left[\frac{12 x^{3}+9}{28}\right]+c$

$=\frac{1}{3}\left(x^{3}-1\right)^{\frac{4}{3}} \cdot \frac{3\left(4 x^{3}+3\right)}{28}+c$

$=\frac{1}{28}\left(x^{3}-1\right)^{\frac{4}{3}}\left(4 x^{3}+3\right)+c$


(ii) $\int \frac{x}{9-4 x^{2}} d x$
Sol :
Let $z=9-4 x^{2}$ then
dz=-4.2xdx=-8xdx

$\Rightarrow-\frac{d z}{8}=x d x$

Now , $\int \frac{x d x}{9-4 x^{2}}=\int \frac{1}{z}\left(-\frac{d z}{8}\right)$

$=-\frac{1}{8} \int \frac{1}{z} d z=\frac{-1}{8} \log z+c$

$=\frac{-1}{8} \log \left(9-4 x^{2}\right)+c$


Question 27
$\int \frac{x^{3}}{\sqrt{x^{2}+3}} d x$
Sol :
Let $z=x^{2}+3$ then dz=2xdx

$\Rightarrow \frac{d z}{2}=x d x$

also , $x^{2}=z-3$

Now , $\int \frac{x^{3}}{\sqrt{x^{2}+3}} d x=\int \frac{x^{2} \cdot x d x}{\sqrt{x^{2}+3}}$

$\int \frac{(z-3) d z}{\sqrt{z}}$

$=\frac{1}{2} \int\left(\frac{z}{\sqrt{z}}-\frac{3}{\sqrt{z}}\right) d z$

$=\frac{1}{2} \int\left(\sqrt{z}-3 z^{-1}\right) d z$

$=\frac{1}{2} \int z^{\frac{1}{2}} d z-\frac{3}{2} \int z^{-\frac{1}{2}} d z$

$=\frac{1}{2} \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}-\frac{3}{2} \cdot 2 z^{\frac{1}{2}}+c$

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

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

$=\sqrt{z} \cdot\left(\frac{z-9}{3}\right)+c$

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

$=\frac{1}{3} \sqrt{x^{3}+3} \cdot\left(x^{3}-6\right)+c$


Question 28
$\int \frac{x^{3}}{\left(x^{2}-a^{2}\right)^{3/ 2}} d x$
Sol :
Let $z=x^{2}-a^{2}$ then dz=2xdx

$\Rightarrow \frac{d 2}{2}=x d x$

also , $z=x^{2}-a^{2}$

$\therefore x^{2}=z+a^{2}$

Now , $\int \frac{x^{3}}{\left(x^{2}-a^{2}\right)^{3 / 2}} d x$

$=\int \frac{x^{2} \cdot x d x}{\left(x^{2}-a^{2}\right)^{3 / 2}}$

$=\int \frac{\left(z+a^{2}\right) }{z^{3 / 2}}\frac{dz}{2}$

$=\frac{1}{2} \int \frac{\left(z+a^{2}\right) d z}{z^{3 / 2}}$

$=\frac{1}{2} \int \frac{z}{z^{3/2}} d z+\frac{1}{2} a^{2} \int \frac{1}{z^{3 / 2}} d z$

$=\frac{1}{2} \int z^{1-\frac{3}{2}} d z+\frac{a^{2}}{2} \int z^{-\frac{3}{2}} d z$

$=\frac{1}{2} \int z^{\frac{-1}{2}} d z+\frac{a^{2}}{2} \int z^{-\frac{3}{2}} dz$

$=\frac{1}{2} \frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{a^{2}}{2} \cdot \frac{z^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+c$

$\frac{1}{2} \cdot 2 z^{\frac{1}{2}}+\frac{a^{2}}{2} \cdot z^{-\frac{1}{2}+1}(-2)+c$

$=z^{\frac{1}{2}}-\frac{a^{2}}{z^{\frac{1}{2}}}+c$

$=\sqrt{z}-\frac{a^{2}}{\sqrt{z}}+c$

$=\frac{z-a^{2}}{\sqrt{z}}+c$

$=\frac{x^{2}-a^{2}-a^{2}}{\sqrt{z}}+c$

$=\frac{x^{2}-2 a^{2}}{\sqrt{x^{2}-a^{2}}}+c$


Question 29
$\int \frac{(\log x)^{2}}{x} d x$
Sol :
Let z=logx then $d z=\frac{1}{x} d x$

Now , $\int \frac{(\log x)^{2}}{x} d x$

$=\int z^{2} d z=\frac{z^{3}}{3}+c$

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


Question 30
(i) $\int \frac{1}{x \log x} d x$
Sol :
Let z=logx then $d z=\frac{1}{x} d x$

Now , $\int \frac{1}{x \log x} d x$

$=\int \frac{1}{\log x \cdot x} d x=\int \frac{1}{z} d z$

=logz+c

=log|logx|+c


(ii) $\int \frac{d x}{x(\log x)^{m}}$
Sol :
Let z=logx then $d z=\frac{1}{x} d x$

Now , $\int \frac{d x}{x(\log x)^{m}}=\int \frac{d z}{z^{m}}=\int z^{-m} d z$

$=\frac{z^{-m+1}}{-m+1}+c=\frac{z^{1-m}}{1-m}+c$

$=\frac{z^{-(m-1)}}{1-m}+c$

$=\frac{1}{z^{(n-1)}} \cdot \frac{1}{1-m}+c$

$=\frac{1}{(1-m) \cdot(\log x)^{m-1}}+c$


Question 31
$\int \frac{d x}{x \sqrt{\log x}}$
Sol :
Let z=logx then $d z=\frac{1}{x} d x$

Now , $\int \frac{d x}{x \cdot \sqrt{\log x}}$

$=\int \frac{1}{\sqrt{\log x}} \cdot \frac{d x}{x}$

$=\int \frac{1}{\sqrt{z}} d z$

$=\int z^{-\frac{1}{2}} d z$

$=\frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c$

$=2 z^{\frac{1}{2}}+c$

$=2 \sqrt{z}+c=2 \sqrt{\log x}+c$

Question 32
(i) $\int \frac{d x}{x(1+\log x)^{2}}$
Sol :
Let z=1+logx then $d z=\frac{1}{x} d x$

Now , $\int \frac{d x}{x \cdot(1+\log x)^{2}}$

$=\int \frac{1}{z^{2}} d z=\int z^{-2} d z$

$=\frac{z^{-2+1}}{-2+1}+c$

$=\frac{z^{-1}}{-1}+c=-\frac{1}{z}+c$

$=-\frac{1}{(1+\log x)}+c$


(ii) $\int \frac{d x}{x+x \log x}$
Sol :
$=\int \frac{d x}{x(1+\log x)}$

Let z=1+logx then $d z=\frac{1}{x} d x$

Now , $\int \frac{d x}{x(1+\log x)}$

$=\int \frac{1}{z} d z$

=log|z|+c

=log|1+logx|+c


Question 33
$\int \frac{\tan x}{\log \cos x} d x$
Sol :
Let z=logcosx then $d z=\frac{1}{\cos x} \times (-\sin x) d x$
dz=-tanxdx
-dz=tanxdx

Now , $\int \frac{\tan x d x}{\log \cos x}=\int \frac{-d z}{z}$

$=-\int \frac{1}{z} d z$

=-logz+c
=-log(logcosx)+c

Question 34
(i) $\int \cot x \cdot \log \sin x d x$
Sol :
Let z=logsinx then $d z=\frac{1}{\sin x} \times \cos x d x$
⇒dz=cotxdx

Now , $\int \cot x \cdot \log \sin x d x$

$=\int \log \sin x \cdot \cot x d x=\int z d z$

$=\frac{z^{2}}{2}+c$

$=\frac{(\log \sin x)^{2}}{2}+c$


(ii) $\int \frac{\cot x}{\log \sin x} d x$
Sol :
Let z=logsinx then $d z=\frac{1}{\sin x} \cdot \cos x d x$  =\cot x dx

Now  , $\int \frac{\cot x d x}{\log \sin x}=\int \frac{1}{z} d z$

=logz+c
=log(logdinx)+c


Question 35
$\int \frac{1}{x \log x \cdot \log (\log x)} d x$
Sol :
Let z=log(logx)dx then $d z=\frac{1}{\log x} \times \frac{1}{x} d x$

$d z=\frac{1}{x(\log x)} d x$

Now , $\int \frac{1}{x \log x \cdot \log (\log x)} d x=\int \frac{d z}{z}$

=logz+c
=log|log(logx)|+c


Question 36
$\int \frac{e^{x}}{1+e^{x}} d x$
Sol :
Let $z=1+e^{x}$ then $d z=e^{x} d x$

Now  , $\int \frac{e^{x} d x}{1+e^{x}}=\int \frac{d z}{z}$

=logz+c
=log(1+e^x)+c


Question 37
$\int \frac{e^{x}}{e^{x}+2} d x$
Sol :
Let $z=e^{x}+2$ then $d z=e^{x} d x$

Now , $\int \frac{e^{x}}{e^{x}+2} d x=\int \frac{d z}{z}$

=log|z|+c

=$\log \left|e^{x}+2\right|+c$


Question 38
$\int \frac{e^{x}}{\left(3-e^{x}\right)^{2}} d x$
Sol :
Let $z=3-e^{x}$ then $d z=-e^{x} d x \Rightarrow-d z=e^{x} d x$

Now , $\int \frac{e^{x}}{\left(3-e^{x}\right)^{2}} d x=\int \frac{-d z}{z^{2}}$

$=-\int z^{-2} d z=-\frac{z^{-2+1}}{-2+1}+c$

$=\frac{-z}{-1}+c=z^{-1}+c$

$=\frac{1}{2}+c$

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



Question 39
$\int e^{\tan x} \sec ^{2} x d x$
Sol :
Let z=tanx then $d z=\sec ^{2} x d x$

Now   , $\int e^{\tan x} \sec ^{2} x d x=\int e^{z} d z$

$=e^{2}+c$

$=e^{\tan x}+c$


Question 40
(i) $\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$
Sol :
Let $z=e^{x}+e^{-x}$ then $d z=\left(e^{x}-e^{-x}\right) d x$

Now , $\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x=\int \frac{d z}{z}$

=logz+c

$=\log \left(e^{x}+e^{-x}\right)+c$


(ii) $\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x$
Sol :
Let $z=e^{2 x}+e^{-2 x}$ then $d z=\left(2 e^{2 x}-2 e^{-2 x}\right) d x$

$dz=2\left(e^{2 x}-e^{-2 x}\right) d x$

$\frac{d z}{2}=\left(e^{2 x}-e^{-2 x}\right) d x$

Now , $\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x=\int \frac{d z}{2 z}$

$=\frac{1}{2} \log z+c$

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


Question 41
(i) $\int x \cdot e^{x^{2}} d x$
Sol :
Let $z=x^{2}$ then dz=2xdx $\Rightarrow \frac{d z}{2}=x d x$

Now , $\int x \cdot e^{x^{2}} d x=\int e^{x^{2}} \cdot x d x$

$=\int e^{2} \cdot \frac{d z}{2}=\frac{1}{2} \int e^{2} d z$

$=\frac{1}{2} \cdot e^{2}+c=\frac{e^{2}}{2}+c$

$=\frac{e^{x^{2}}}{2}+c$


(ii) $\int \frac{x}{e^{x^{2}}} d x$
Sol :
$\int \frac{x}{e^{x^{2}}} d x=\int x \cdot e^{-x^{2}} d x$

Let $z=-x^{2}$ then dz=-2xdx $\Rightarrow-\frac{d z}{2}=x d x$

Now , $\int x \cdot e^{-x^{2}} d x=\int e^{-x^{2}} x d x$

$=\int e^{z}\left(-\frac{d z}{2}\right)=-\frac{1}{2} \int e^{2} d z$

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


Question 42
$\int \frac{\left(\tan ^{-1} x\right)^{2}}{1+x^{2}} d x$
Sol :
Let $z=\tan ^{-1} x$ then $d z=\frac{1}{1+x^{2}} d x$

Now , $\int \frac{\left(\tan ^{-1} x\right)^{2}}{1+x^{2}} d x$

$=\int z^{2} d z=\frac{z^{3}}{3}+c$

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



Question 43
$\int \frac{e^{x} \sin x}{e^{x}+\cos x} d x$
Sol :
Let $z=e^{x}+\cos x$ then $d z=\left(e^{x}-\sin x\right) d x$

Now , $\int \frac{d z}{z}=\log |z|+c$

$=|\log | e^{x}+\cos x |+c$


Question 44
$\int \frac{d x}{\left(1+x^{2}\right) \cdot \tan ^{-1} x}$
Sol :
Let $z=\tan ^{-1} x$ then $d z=\frac{1}{1+x^{2}} d x$

Now  , $\int \frac{d x}{\left(1+x^{2}\right) \cdot \tan ^{-1} x}=\int \frac{d z}{z}$

$=\int \frac{1}{z} d z$

=log|z|+c
$=\log \left(\tan ^{-1} x\right)+c$


Question 45
$\int \frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}} d x$
Sol :
Let $z=\sin ^{-1} x$ then $d z=\frac{1}{\sqrt{1-x^{2}}} d x$

Now , $\int \frac{e^{\sin ^{-1}x}}{\sqrt{1-x^{2}}} d x=\int e^{z} d z$

$=e^{z}+c=e^{\sin ^{-1} x}+c$


Question 46
$\int \frac{1}{\sqrt{1-x^{2}}} \frac{1}{\cos ^{-1} x} d x$
Sol :
Let $z=\cos ^{-1} x$ then $d z=-\frac{1}{\sqrt{1-x^{2}}} d x$

Now , $\int \frac{1}{\cos ^{-1} x}=\frac{d x}{\sqrt{1-x^{2}}}$

$=\int \frac{1}{z}(-d z)$

$=-\int \frac{1}{z} d z$

=-log|z|+c
=$=-\log \left|\cos ^{-1} x\right|+c$


Question 47
$\int \frac{x^{2} \cdot \tan ^{-1}\left(x^{3}\right)}{1+x^{6}} d x$
Sol :
Let $z=\tan ^{-1}\left(x^{3}\right)$ then

$d z=\frac{1 \times 3 x^{2}}{1+\left(x^{3}\right)^{2}} d x=\frac{3 x^{2}}{1+x^{6}} d x$

$\frac{d z}{3}=\frac{x^{2}}{1+x^{6}} d x$

Now , $\int \frac{x^{2} \cdot \tan ^{-1}\left(x^{3}\right)}{1+x^{6}} d x$

$=\int \frac{d z}{3} \cdot z=\frac{1}{3} \int z d z$

$=\frac{1}{3} \cdot \frac{z^{2}}{2}+c=\frac{1}{6} z^{2}+c$

$=\frac{1}{6}\left[\tan ^{-1}\left(x^{3}\right)\right]^{2}+c$



Question 48
$\int(2 x+4) \cdot \sqrt{x^{2}+4 x+3} d x$
Sol :
Let $z=x^{2}+4 x+3$ then dz=(2x+4)dx

Now  , $\int(2 x+4) \sqrt{x^{2}+4 x+3} d x$

$=\int \sqrt{x^{2}+4 x+3} \cdot(2 x+4) d x$

$=\int \sqrt{z} \cdot d z=\int z^{\frac{1}{2}} d z$

$=\frac{z^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=\frac{2}{3} \cdot z^{\frac{3}{2}}+c$

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


Question 49
$\int \frac{1+x^{2}}{x^{2}} \cdot e^{x-\frac{1}{x}} d x$
Sol :
Let $z=x-\frac{1}{x}$ then $d z=\left[1-\left(\frac{-1}{x^{2}}\right)\right]$

$=\left[\frac{1+x^{2}}{x^{2}}\right] d x$

Now , $\int \frac{1+x^{2}}{x^{2}} \cdot e^{x-\frac{1}{x}} d x$

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

$=\int e^{z} d z$

$=e^{2}+c$

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



Question 50
$\int \frac{d x}{x+\sqrt{x}}$
Sol :
Let $z=1+\sqrt{x}$ then $d z=\frac{1}{2 \sqrt{x}} d x \quad \Rightarrow 2 d z=\frac{1}{\sqrt{x}} d x$

Now , $\int \frac{d x}{x+\sqrt{x}}=\int \frac{d x}{\sqrt{x}(\sqrt{x}+1)}$

$=\int \frac{2 d z}{2}=2 \int \frac{1}{2} d z$

=2logz+c

=2log(1+√x)+c


Question 51
$\int \frac{x}{\left(x^{2}+2\right)^{1 / 3}} d x$
Sol :
Let $z=x^{2}+2$ then dz=2xdx $\Rightarrow \frac{d z}{2}=x d x$

Now , $\int \frac{x}{\left(x^{2}+2\right)^{1/3}} d x=\int \frac{1}{z^{1/3}} \cdot \frac{d z}{2}$

$=\frac{1}{2} \int z^{\frac{-1}{3}} d z$

$=\frac{1}{2} \cdot \frac{z^{-\frac{1}{3}+1}}{\frac{-1}{3}+1}+c$

$=\frac{1}{2} \times \frac{3}{2} \times z^{\frac{2}{3}}+c$

$=\frac{3}{4} z^{\frac{2}{3}}+c$

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


Question 52
$\int \frac{\tan ^{2} x \cdot \sec ^{2} x d x}{1+\tan ^{6} x}$
Sol :
Let z=tanx then $d z=\sec ^{2} x d x$

Now , $\int \frac{\tan ^{2} x \cdot \sec ^{2} x}{1+\tan ^{6} x}=\int \frac{z^{2}}{1+z^{6}} d z$

$=\int \frac{z^{2}}{1+\left(z^{3}\right)^{2}} d z$

Let $y=z^{3}$ then

$d y=3z^{2} d z \Rightarrow \frac{d y}{3}=z^{2} d z$

Now , $\int \frac{z^{2}}{1+\left(z^{3}\right)^{2}} d z=\int \frac{1}{1+y^{2}} \cdot \frac{d y}{3}$

$=\frac{1}{3} \int \frac{1}{1+y^{2}} d y=\frac{1}{3} \tan ^{-1} y+c$

$=\frac{1}{3} \tan ^{-1}\left(z^{3}\right)+c$

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


Question 53
$\int \frac{d x}{x \log x[\log (\log x)]}$
Sol :
Let z=(log x) then $d z=\frac{1}{\log x}\times \frac{x}{x} d x$
$d z=\frac{d x}{x \cdot \log x}$

Now , $\int \frac{d x}{x \log x \cdot[\log (\log x)]}=\int \frac{d z}{z}$

=log z+c

=log|log(logx)|+c


Question 54
$\int \frac{e^{2 x}}{\sqrt[4]{e^{x}+1}} d x$
Sol :
Let $z=e^{x}+1$ then $dz=e^{x} d x$

Now , $\int \frac{e^{2 x}}{\sqrt[4]{e^{x}+1}} d x=\int \frac{e^{x} \cdot e^{x}}{\sqrt[4]{e^{x}+1}} d x$

$=\int \frac{(z-1) d z}{\sqrt[4]{z}}$

$=\int \frac{z-1}{z^{1 / 4}} d z$

$=\int \frac{z}{z^{1/4}} d z-\int \frac{1}{z^{1 /4}} d z$

$=\int z^{1-\frac{1}{4}} d z-\int z^{-\frac{1}{4}} d z$

$=\int z^{\frac{3}{4}} d z-\int z^{-\frac{1}{4}} d z$

$=\frac{z^{\frac{3}{4}+1}}{\frac{3}{4}+1}-\frac{z^{-\frac{1}{4}+1}}{\frac{-1}{4}+1}+c$

$=\frac{z^{\frac{7}{4}}}{\frac{7}{4}}-\frac{z^{\frac{3}{4}}}{\frac{3}{4}}+c$

$=\frac{4}{7} \cdot z^{\frac{7}{4}}-\frac{4}{3} \cdot z^{\frac{3}{4}}+c$

$=\frac{4}{7}\left(e^{x}+1\right)^{\frac{7}{4}}-\frac{4}{3}\left(e^{x}+1\right)^{\frac{3}{4}}+c$

$=\left(e^{x}+1\right)^{\frac{3}{4}}\left[\frac{4}{7}\left(e^{x}+1\right)-\frac{4}{3}\right]+c$

$=\left(e^{x}+1\right)^{\frac{3}{4}}\left[\frac{4 e^{x}+4}{7}-\frac{4}{3}\right]+c$

$=\left(e^{x}+1\right)^{\frac{3}{4}}\left[\frac{12 e^{x}+12-28}{21}\right]+c$

$=\left(e^{x}+1\right)^{\frac{3}{4}}\left[\frac{12 e^{x}-16}{21}\right]+c$

$=\frac{4}{21}\left(e^{x}+1\right)^{\frac{3}{4}}\left[3 e^{x}-4\right]+c$


Question 55
$\int x \cdot \sin x^{2} d x$
Sol :
Let $z=x^{2}$ then dz=2xdx
$\Rightarrow \frac{d z}{2}=x d x$

Now , $\int x \cdot \sin x^{2} d x=\int \sin x^{2} \cdot x d x$

$=\int \sin z \cdot \frac{d z}{z}$

$=\frac{1}{z} \int \sin z d z$

$=\frac{1}{2}(-\cos z)+c$

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


Question 57
$\int \frac{x^{3}}{\left(1+x^{2}\right)^{2}} d x$
Sol :
Let $z=1+x^{2}$ then dz=2xdx $\Rightarrow \frac{d z}{2}=x d x$

Also , $z=1+x^{2}$  $\therefore x^{2}=z-1$

Now ,$\int \frac{x^{3}}{\left(1+x^{2}\right)^{2}} d x=\int \frac{x^{2} \cdot x d x}{\left(1+x^{2}\right)^{2}}$

$=\int \frac{(z-1)}{z^{2}} \frac{d z}{2}$

$=\frac{1}{2} \int \frac{(z-1)}{z^{2}} d z$

$=\frac{1}{2} \int \frac{z}{z^{2}} d z-\frac{1}{2} \int \frac{1}{z^{2}} d z$

$=\frac{1}{2} \int \frac{1}{z} d z-\frac{1}{2} \int z^{-2} d z$

$=\frac{1}{2} \log z-\frac{1}{2} \frac{z^{-2+1}}{-2+1}+c$

$\frac{1}{2} \log z-\frac{1}{2} \frac{z^{-1}}{-1}+c$

$=\frac{1}{2} \log z+\frac{1}{2 z}+c$

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


Question 58
$\int x^{2} \cdot\left(1 \cdot x^{2}\right)^{\frac{5}{2}} d x$
Sol :
Let  $z=1-x^{2}$ then dz=-2xdx $\Rightarrow \frac{-d z}{2}=x d x$

Also , $x^{2}=1-z$

Now , $\int x^{3} \cdot\left(1-x^{2}\right)^{\frac{5}{2}} d x$

$=\int\left(1-x^{2}\right)^{\frac{5}{2}} x^{2} \cdot x d x$

$=\int(z)^{\frac{5}{2}} \cdot(1-z)\left(-\frac{d z}{2}\right)$

$=-\frac{1}{2} \int z^{\frac{5}{2}}(1-z) d z$

$=-\frac{1}{2} \int z^{\frac{5}{2}} d z+\frac{1}{2} \int z^{\frac{5}{2}} \cdot z d z$

$=-\frac{1}{2} \int z^{\frac{5}{2}} d z+\frac{1}{2} \int z^{\frac{7}{2}} d z$

$=-\frac{1}{2} \frac{z^{\frac{5}{2}+1}}{\frac{5}{2}+1}+\frac{1}{2} \frac{z^{\frac{9}{2}}}{\frac{9}{2}}+c$

$=\frac{-1}{2} \frac{z^{\frac{7}{2}}}{\frac{7}{2}}+\frac{1}{2} \frac{z^{\frac{9}{2}}}{\frac{9}{2}}+c$

$=-\frac{1}{2} \times \frac{2}{7} \cdot z^{\frac{7}{2}}+\frac{1}{2} .\frac{2}{9} z^{\frac{9}{2}}+c$

$=\frac{-1}{7} z^{\frac{7}{2}}+\frac{1}{9} z^{\frac{9}{2}}+c$

$=\frac{1}{9}\left(1-x^{2}\right)^{\frac{9}{2}}-\frac{1}{7}\left(1-x^{2}\right)^{\frac{7}{2}}+c$


Question 59
$\int \sqrt{\sec x+1} d x$
Sol :
$=\int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{1-2 \sin ^{2} \frac{x}{2}}} d x$

Let $z=\sqrt{2} \sin \frac{x}{2}$ then $d z=\frac{1}{2} \sqrt{2} \cdot \cos \frac{x}{2} d x$

Now  ,  $\int \frac{\sqrt{2} \cos \frac{x}{2} d x}{\sqrt{1-(\sqrt{2} \sin \frac{x}{2})^{2}}}$

$=\int \frac{2 d z}{\sqrt{1-z^{2}}}$

$=2 \int \frac{1}{\sqrt{1-z^{2}}} d z$

$=2 \sin ^{-1} z+c$

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


Question 60
$\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$
Sol :
$=\int \frac{2 \sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x$

Dividing numerator and denominator by cos⁴x

$=\int \frac{\frac{2 \sin x \cdot \cos x}{\cos ^{4} x}}{\frac{\sin ^{4} x+\cos ^{4} x}{\cos ^{4} x}} d x$

$=\int \frac{2 \sin x}{\frac{\cos ^{2} x \cdot \cos x}{\tan^{4} x+1}} d x$

$=\int \frac{2 \tan x \cdot \sec ^{2} x d x}{1+\tan ^{4} x} d x$

$=\int \frac{2 \tan x}{1+\tan ^{4} x} \cdot \sec ^{2} x d x$

Let z=tanx then $d z=\sec ^{2} x d x$

Now , $\int \frac{2 \tan x}{1+\tan ^{4} x} \cdot \sec ^{2} x d x$

$=\int \frac{2 z}{1+z^{4}} d z$

Let $y=z^{2}$ then dy=2zdz

Now , $\int \frac{2zd z}{1+\left(z^{2}\right)^{2}}=\int \frac{d y}{1+y^{2}}$

$=\tan ^{-1} y+c$

$=\tan ^{-1}\left(z^{2}\right)+c$

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


Question 61
(i) $\int \frac{x^{e-1}-e^{x-1}}{x^{e}-e^{x}} d x$
Sol :
Let $z=x^{2}-e^{x}$ then $d z=\left(e \cdot x^{e-1}-e^{x}\right) d x$
$d z=\left(e x^{e-1}-\frac{e}{e} \cdot e^{x}\right) d x$
$d z=e\left(x^{e-1}-\frac{1}{e} \cdot e^{x}\right) d x$

[∵ x^m.xⁿ=x^m+n]

$\frac{d z}{e}=\left(x^{e-1}-e^{x-1}\right) d x$

$\int \frac{\left(x^{e-1}-e^{x-1}\right) d x}{x^{e}-e^{x}}=\int \frac{d z}{e-z}$

$=\frac{1}{e} \int \frac{1}{z} d z=\frac{1}{e} \log z+c$

$=\frac{1}{e} \log \left|x^{e}-e^{x}\right|+c$


(ii) $\int \frac{1+\cos \theta}{\theta+\sin \theta} d \theta$
Sol :
Let z=θ+sinθ then dz=(1+cosθ)dθ

Now , $\int \frac{1+\cos \theta}{\theta+\sin \theta}d \theta=\int \frac{d z}{z}$

=log z+c

=log|θ+sinθ|+c


Question 62
(i) $\int \frac{1+\tan x}{x+\log \sec x} d x$
Sol :
Let z=x+logsecx then $d z=\left(1+\frac{1 \times \sec x}{\sec x} \cdot \tan x\right) d x$

dz=(1+tanx)dx

Now  , $\int \frac{1+\tan x}{x+\log \sec x} d x=\int \frac{d z}{z}$

=log z+c
=log|x+logsecx|+c


(ii) $\int \frac{\left(3 \tan ^{2} x+2\right) \cdot \sec ^{2} x}{\tan ^{3} x+2 \tan x+5} d x$
Sol :
Let $z=\tan ^{3} x+2 \tan x+5$ then $d z=\left(3 \tan ^{2} x \cdot \sec ^{2} x+2 \sec ^{2} x\right) d x$
$d z=\sec ^{2} x\left(3 \tan ^{2} x+2\right) d x$
$d z=\left(3 \tan ^{2} x+2\right) \sec ^{2} x d x$

Now , $\int \frac{\left(3 \tan ^{2} x+2\right) \sec ^{2} x}{\tan ^{2} x+2 \tan x+5}=\int \frac{d z}{z}$

=log z+c

=$\log \left|\tan ^{3} x+2 \tan x+5\right|+c$

(iii) $\int \frac{1+\sin 2 x}{x+\sin ^{2} x} d x$
Sol :
Let $z=x+\sin ^{2} x$ then dz=(1+2sinx.cosx)dx
dz=(1+sin2x)dx

Now  , $\int \frac{1+\sin 2 x}{x+\sin ^{2} x} d x=\int \frac{d z}{z}$

=logz+c
=log|x+sin²x|+c


(iv) $\int \frac{d x}{x-\sqrt{x}}$
Sol :
$=\int \frac{d x}{\sqrt{x} \cdot \sqrt{x}-\sqrt{x}}=\int \frac{d x}{\sqrt{x}(\sqrt{x}-1)}$

Let  $z=\sqrt{x}-1$ then

$d z=\frac{1}{2 \sqrt{x}} \Rightarrow 2 d z=\frac{1}{\sqrt{x}} d x$

Now  , $\int \frac{d x}{x-\sqrt{x}}=\int \frac{d x}{\sqrt{x}(\sqrt{x}-1)}$

$=\int \frac{2 d z}{z}$ =2logz+c

$=2 \log |\sqrt{x}-1|+c$


Question 63
(i) $\int \frac{\tan x}{\sec x+\cos x} d x$
Sol :
$\int \frac{\tan x}{\sec x+\cos x} d x=\int \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\cos x} d x$

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

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

Let z=cosx then dz=-sinxdx
-dz=sinxdx

Now  , $\int \frac{\tan x}{\sec x+\cos x}=\int \frac{\sin x}{1+\cos ^{2} x} d x$

$=\int \frac{-d z}{1+z^{2}}=-\int \frac{1}{1+z^{2}} d z$

=$-\tan ^{-1} z+c$

=$-\tan ^{-1}(\cos x)+c$


(ii) $\int \frac{1-\cot x}{1+\cot x} d x$
Sol :
$=\int \frac{1-\frac{\cos x}{\sin x}}{1+\frac{\cos x}{\sin x}} d x$

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

$=\int \frac{\sin x-\cos x}{\sin x+\cos x} d x$

Let z=sinx+cosx then dz=(cosx-sinx)dx
dz=-(-cosx+sinx)dx
-dz=(sinx-cosx)dx

Now , $\int \frac{1-\cot x}{1+\cot x} d x=\int \frac{\sin x-\cos x}{\sin x+\cos x}d x$

$=\int \frac{-d z}{z}$

=-log z+c

=-log|sinx+cosx|+c


Question 64
$\int \frac{e^{\tan ^{-1} x}}{1+x^{2}} d x$
Sol :
Let $z=\tan ^{-1} x$ then $d z=\frac{1}{1+x^{2}} d x$

Now ,$\int \frac{e^{\tan ^{-1}x}}{1+x^{2}} d x=\int e^{z} d z$
$=e^{z}+c$

$=e^{t a n ^{-1}}+c$


Question 65
$\int \sec x \cdot \log (\sec x+\tan x) d x$
Sol :
Let z=log(secx+tanx) then

$d z=\frac{1\times \sec x \tan x+\sec ^{2} x }{\sec x+\tan x}dx$

$d z=\frac{\sec x(\sec x+\tan x)}{(\sec x+\tan x)}dx$

dz=secxdx

Now , $\int \sec x \cdot \log (\sec x+\tan x) d x$

$=\int \log (\sec x+\tan x) \cdot \sec x d x$

$=\int z d z=\frac{z^{2}}{2}+c$

$=\frac{1}{2}[\log (\sec x+\tan x)]^{2}+c$


Question 66
(i) $\int \frac{\tan ^{4} \sqrt{x} \cdot \sec ^{2} \sqrt{x}}{\sqrt{x}} d x$
Sol :
Let z=tan√x then $d z=\sec ^{2} \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} d x$

$2 d z=\frac{\sec ^{2} \sqrt{x}}{\sqrt{x}} d x$

Now , $\int \frac{\tan ^{4} \sqrt{x} \cdot \sec ^{2} \sqrt{x}}{\sqrt{x}} d x=\int z^{4} 2 d z$

$=2 \int z^{4} d z=\frac{2 z^{5}}{5}+c$

$=\frac{2}{5}\left(\tan ^{5} \sqrt{x}\right)+c$


(ii) $\int \frac{\tan x \cdot \sec ^{2} x}{1-\tan ^{2} x} d x$
Sol :
Let z=tanx then $d z=\sec ^{2} x d x$

Now , $\int \frac{z}{1-z^{2}} d z$

Let $y=1-z^{2}$ then dy=-2zdz $\Rightarrow \frac{d y}{2}=zd z$

Now , $\int \frac{1}{1-z^{2}}z d z=\int \frac{1}{y}\left(-\frac{d y}{2}\right)$

$=\frac{-1}{2} \int \frac{1}{y} d y=\frac{-1}{2} \log y+c$

$=-\frac{1}{2} \log \left(1-z^{2}\right)+c$

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



Question 67
(i) $\int \frac{d x}{1+\tan x}$
Sol :
$=\int \frac{d x}{1+\frac{\sin x}{\cos x}}=\int \frac{\cos x}{\sin x+\cos x} d x$

$=\int \frac{2 \cos x}{2(\sin x+\cos x)} d x$

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

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

$=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{d z}{2}$

where z=sinx+cosx
dz=(cosx-sinx)dx

Now , $\frac{1}{2} \int d x+\frac{1}{2} \int \frac{1}{2} d z$

$=\frac{1}{2} x+\frac{1}{2} \log z+c$

$\frac{x}{2}+\frac{1}{2} \log |\sin x+\cos x|+c$


(ii) $\int \frac{d x}{1-\tan x}$
Sol :
$=\int \frac{\cos x}{\cos x-\sin x} d x=\int \frac{2 \cos x}{2(\cos x-\sin x)} d x$

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

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

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

Let z=cosx-sinx then dz=(-sinx-cosx)dx
dz=-(cosx+sinx)dx
-dz=(cosx+sinx)dx

Now , $\frac{1}{2} \int d x+\frac{1}{2} \int \frac{-d z}{z}$

$=\frac{x}{2}-\frac{1}{2} \log z+c$

$=\frac{x}{2}-\frac{1}{2} \log (\cos x-\sin x)+c$


Question 68
$\int \frac{a}{b+c e^{x}} d x$
Sol :
$=\int \frac{a}{e^{x}\left(\frac{b}{e^{x}}+c\right)} d x$

$=\int \frac{a}{e^{x}\left(b \cdot e^{-x}+c\right)} d x$

$=\int \frac{a \cdot e^{-x} d x}{\left(b \cdot e^{-x}+c\right)}$

Let $z=b \cdot e^{-x}+c \Rightarrow d z=b e^{-x} \times-1 d x$

$-\frac{d z}{b}=e^{-x} d x$

Now , $\int \frac{a \cdot e^{-x} d x}{b \cdot e^{-x}+c}=a \int \frac{-d z}{b \cdot z}$

$=-\frac{a}{b} \int \frac{1}{z} d z$

$=-\frac{a}{b} \log z+k$

$=-\frac{a}{b} \log \left(\frac{b}{e^{x}}+c\right)+K$

$=-\frac{a}{b} \log \left(\frac{b+c e^{x}}{e^{x}}\right)+k$

$\log \frac{m}{n}=\log m-\log n$

$=-\frac{a}{b} \log \left(b+c e^{x}\right)+\frac{a}{b} \log e^{x}$

$=-\frac{a}{b} \log \left(b+c e^{x}\right)+\frac{a}{b} x$

$=\frac{a}{b} x-\frac{a}{b} \log \left(b+c e^{x}\right)+k$

$=\frac{9}{6}\left[x-\log \left(b+c e^{x}\right)\right]+k$


Question 69
$\int \frac{e^{x}}{e^{x}+e^{-x}} d x$
Sol :
$=\frac{1}{2} \int \frac{2 e^{x}}{e^{x}+e^{-x}} d x$

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

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

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

$=\frac{x}{2}+\frac{1}{2} \int \frac{d z}{z}$

$=\frac{x}{2}+\frac{1}{2} \log z+c$

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

$=\frac{1}{2}\left[x+\log \left(e^{x}+e^{-x}\right)\right]+c$


Question 70
$\int \frac{(x+1)(x+\log x)^{2}}{x} d x$
Sol :
$\int \frac{(x+1)(x+\log x)^{2}}{x} d x=\int \frac{(x+\log x)^{2}(x+1)}{x} d x$

Let z=(x+logx)

then $dz=\left(1+\frac{1}{x}\right) d x$

$dz=\left(\frac{1+x}{x}\right) d x$

Now ,$\int \frac{(x+1)}{x}(x+\log x)^{2} d x$

$=\int z^{2} d z=\frac{z^{3}}{3}+c$

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


Question 71
$\int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x$
Sol :
$=\int \frac{2 \sin x \cdot \cos x}{(a+b \cos x)^{2}} d x$

$=\int \frac{2 \cos x \sin x d x}{(a+b \cos x)^{2}}$

$=2 \int \frac{\cos x \cdot \sin x d x}{(a+b \cos x)^{2}}$

Let z=a+bcosx

$\cos x=\frac{z-a}{b}$

then $d z=-b(\sin x) d x$

$\frac{d z}{-b}=\sin x d x$

Now ,  $2 \int \frac{\cos x \cdot \sin x d x}{(a+b \cos x)^{2}}$

$=2 \int \frac{z-a}{b \cdot z^{2}} \cdot\left(\frac{-d z}{b}\right)$

$=-\frac{2}{b^{2}} \int \frac{z-a}{z^{2}} d z$

$=-\frac{2}{b^{2}} \int \frac{z}{z^{2}} d z+\frac{2}{b^{2}} \int \frac{a}{z^{2}} d z$

$=\frac{-2}{b^{2}} \int \frac{1}{z} d z+\frac{2 a}{b^{2}} \int z^{-2} d z$

$=-\frac{2}{b^{2}} \log z+\frac{2 a}{b^{2}} \cdot \frac{z^{-2+1}}{-2+1}+c$

$=\frac{-2}{b^{2}} \log z+\frac{2 a}{b^{2}} \frac{z^{-1}}{-1}+c$

$=-\frac{2}{b^{2}} \log z-\frac{2a}{b^{2}} \cdot \frac{1}{z}+c$

$=\frac{-2}{b^{2}}\left[\log z+\frac{a}{z}\right]+c$

$=-\frac{2}{b^{2}}\left[\log (a+b \cos x)+\frac{a}{(a+b \cos x)}\right]+c$


Question 72
$\int \frac{10 \cdot x^{3}+10^{x} \cdot \log 10 d x}{10^{x}+x^{10}}$
Sol :
Let $z=10^{x}+x^{10}$ then $d z=10^{x} \cdot \log 10+10 \cdot x^{10-1} d x$

$d z=\left(10 \cdot x^{9}+10^{x} \cdot \log _{e} 10\right) d x$

Now , $\int \frac{10 \cdot x^{9}+10^{x} \cdot \log 10}{10^{x}+x^{10}} d x=\int \frac{d z}{z}$

=log z+c

$=\log \left(10^{x}+x^{10}\right)+c$


Question 73
(i) $\int \frac{x^{5}}{\sqrt{1+x^{2}}} d x$
Sol :
Let $z=1+x^{2}$⇒$x^{2}=z-1$ then

dz=2xdx ⇒ $\frac{d z}{2}=x d x$

$=\frac{1}{2} \int \frac{(z-1)^{2}}{\sqrt{z}} d z$

$=\frac{1}{2} \int \frac{z^{2}+1-2 z}{\sqrt{z}} d z$

$\frac{1}{2}\left[\int \frac{z^{2}}{\sqrt{z}} d z+\int \frac{1}{\sqrt{z}} d z-2 \int \frac{z}{\sqrt{z}} d z\right]$

$=\frac{1}{2}\left[\int z^{2-\frac{1}{2}} d z+\int z^{-\frac{1}{2}} d z-2 \int z^{\frac{1}{2}} d z\right]+c$

$=\frac{1}{2}\left[\int z^{\frac{3}{2}} d z+\int z^{-\frac{1}{2}} d z-2 \int z^{\frac{1}{2}} d z\right]+c$

$=\frac{1}{2}\left[\frac{2}{5} \cdot 2^{\frac{5}{2}}+2 z^{\frac{1}{2}}-2 \cdot \frac{2}{3} \cdot z^{\frac{3}{2}}\right]+c$

$\frac{1}{5} z^{\frac{5}{2}}-\frac{2}{3} z^{\frac{3}{2}}+z^{\frac{1}{2}}+c$

$\frac{1}{5}\left(1+x^{2}\right)^{\frac{5}{2}}-\frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}+\sqrt{1+x^{2}}+c$


(ii) $\int \frac{x^{3}}{\sqrt{1-x^{8}}} d x$
Sol :
$=\int \frac{x^{3}}{\sqrt{1-\left(x^{4}\right)^{2}}} d x$

Let $z=x^{4}$ then

$d z=4 x^{3} d x \Rightarrow \frac{d z}{4}=x^{3} d x$

Now , $\int \frac{x^{3}}{\sqrt{1-x^{2}}} d x=\int \frac{d z}{4 \cdot \sqrt{1-z^{2}}}$

$=\frac{1}{4} \int \frac{1}{\sqrt{1-z^{2}}} d z$

$=\frac{1}{4} \sin ^{-1} z+c$

$=\frac{1}{4} \sin ^{-1} x^{4}+c$


(iii) $\int e^{3 \log x}\left(1+x^{4}\right)^{-1} d x$
Sol :
$\int e^{3 \log x}\left(1+x^{4}\right)^{-1} d x$

$=\int \frac{e^{\log x^{3}}}{\left(1+x^{4}\right)} d x$

$=\int \frac{x^{3} d x}{1+x^{4}}$

Let $z=1+x^{4}$ then

$d z=4 x^{3} d x \Rightarrow \frac{d z}{4}=x^{3} d x$

Now , $\int \frac{x^{3}}{1+ x^{4}} d x=\int \frac{d z}{4 \cdot z}$

$=\frac{1}{4} \int \frac{1}{z} d z=\frac{1}{4} \log z+c$

$=\frac{1}{4} \log \left(1+x^{4}\right)+c$


Question 74
$\int \frac{d x}{\sqrt[3]{(x+1)^{2} \cdot(x-1)^{4}}}$
Sol :
$=\int \frac{d x}{\sqrt[3]{(x+1)^{2} \cdot(x-1)^{2}(x-1)^{2}}}$

$=\int \frac{d x}{\sqrt[3]{\left(\frac{x+1}{1-x}\right)^{2}-(1-x)^{6}}}$

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

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

Let $z=\frac{1+x}{1-x}$ then

$dz=\frac{(1-x) \cdot 1-(1+x)(-1)}{(1-x)^{2}} d x$

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

$\dfrac{d z}{2}=\frac{1}{(1-x)^{2}} d x$

Now , $\frac{1}{2} \int \frac{d z}{z^{\frac{2}{3}} }$

$=\frac{1}{2} \int z^{-\frac{2}{3}} d z$

$=\frac{1}{2} \cdot \frac{z^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+c$

$=\frac{3}{2} \cdot z^{\frac{1}{3}}+c$

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

$=\frac{3}{2} \cdot \sqrt[3]{\frac{1+x}{1-x}}+c$


Question 75
$\int \frac{d x}{\sqrt{1+x^{2}} \cdot \sqrt{\log (x+\sqrt{1+x^{2}})}}$
Sol :
Let $z=\log (x+\sqrt{1+x^{2}})$ then

$d z=\frac{1+\frac{x}{\sqrt{1+x^{2}}} d x}{x+\sqrt{1+x^{2}}}$

$d z=\frac{\frac{x+\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}}}{x+\sqrt{1+x^{2}}} d x$

$d z=\frac{1}{\sqrt{1+x^{2}}} d x$

Now , $\int \frac{d z}{\sqrt{z}}=\int z^{-\frac{1}{2}} d z$

$=\frac{z^{-1+1}}{-\frac{1}{2}+1}+c=2 \cdot z^{\frac{1}{2}}+c$

$=2 \sqrt{z}+c$

$=2 \sqrt{\log (x+\sqrt{1+x^{2}})}+c$



Question 76
$\int[F(a x+b)]^{n} \cdot f^{\prime}(a x+b) d x$
Sol :
Let z=[F(ax+b)] then $d z=f^{\prime}(a x+b) x a d x$

$\frac{d z}{a}=f^{\prime}(a x+b) d x$

Now , $\int[f(a x+b)]^{n} \cdot f^{\prime}(a x+b) d x$

$=\int z^{n} \frac{d z}{a}$

$=\frac{1}{a} \cdot \frac{z^{n+1}}{n+1}+c$

$=\frac{1}{a} \cdot \frac{[F(a x+b)]^{n+1}}{n+1}$

$=\frac{[F(a x+b)]^{n+1}}{(n+1) a}$


Question 77
$\int \frac{d x}{\sqrt{\sin ^{3} x \cdot \sin (x+\alpha)}}$
Sol :
$\int \frac{d x}{\sqrt{\sin ^{3} x \cdot(\sin x \cdot \cos \alpha+\cos x \cdot \sin \alpha)}}$

$=\int \frac{d x}{\sqrt{\sin ^{4} x \cdot \cos \alpha+\sin ^{3} x \cdot \cos x \cdot \sin \alpha}}$

$=\int \frac{d x}{\sqrt{\sin ^{2} x\left(\sin ^{2} x \cdot \cos \alpha+\sin x \cdot \cos x \cdot \sin \alpha\right)}}$

$=\int \frac{d x}{\sin x \cdot \sqrt{\sin ^{2} x \cdot \cos \alpha+ \frac{\sin ^{2} x \cdot \cos x \cdot \sin a}{\sin x}}}$

$=\int \frac{d x}{\sin x\sqrt{\sin ^{2} x(\cos \alpha+\cot x \cdot \sin \alpha)}}$

$=\int \frac{d x}{\sin x \cdot \sin x \sqrt{\cos ^{2} \alpha+\cot x.sin \alpha}}$

$=\int \frac{d x}{\sin x \cdot \sin x \sqrt{\cos \alpha+\cot x \cdot \sin \alpha }}$

$=\int \frac{\operatorname{cosec}^{2} x d x}{\sqrt{\cos \alpha+\sin \alpha \cdot \cot x}}$

Let z=cosɑ +sinɑ.cotx then dz=sinɑ.(-cosec²x)dx

$\frac{d z}{\sin \alpha}=\operatorname{cosec}^{2} x d x$

Now , $\int \frac{d x}{\sqrt{\sin ^{3} x \cdot \sin (x+\alpha)}}$

$=\int \frac{\operatorname{cosec}^{2} x d x}{\sqrt{\cos \alpha+\sin \alpha \cdot \cot x}}$

$=\int \frac{-d z}{\sin \alpha \sqrt{2}}$

$=\frac{-1}{\sin \alpha} \int z^{-\frac{1}{2}} d z$

$=\frac{-1}{\sin \alpha} \cdot \frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c$

$=-2 \operatorname{cosec} \alpha z^{\frac{1}{2}}+c$

$=-2 \operatorname{cosec} \alpha \sqrt{z}+c$

$=-2cosec \alpha \sqrt{\cos \alpha+\sin \alpha \cdot \cot x}+c$