Indefinite Integration

$\begin{aligned} \text { Let } I & =\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{5 / 4}} \\ & =\int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3 / 4}(x-2)^2} \end{aligned}$

On putting, $\frac{x+1}{x-2}=t \Rightarrow \frac{d}{d x}\left(\frac{x+1}{x-2}\right)=\frac{d t}{d x}$

$\begin{aligned} & \Rightarrow \quad \frac{(x-2) \frac{d}{d x}(x+1)-(x+1) \frac{d}{d x}(x-2)}{(x-2)^2}=\frac{d t}{d x} \\ & \Rightarrow \quad \frac{(x-2)(1)-(x+1)(1)}{(x-2)^2}=\frac{d t}{d x} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{-3}{(x-2)^2} d x=d t \\ & \text { So, } I=\int \frac{d t}{-3 t^{3 / 4}}=-\frac{1}{3}\left(\frac{t^{-3 / 4+1}}{-\frac{3}{4}+1}\right)+C \\ & =-\frac{1}{3}\left(4 t^{1 / 4}\right)+C=-\frac{4}{3}\left(\frac{x+1}{x-2}\right)^{1 / 4}+C \\ & \end{aligned}$

$\text { We have, } f(x)=\int \frac{\sqrt{x}}{(1+x)^2} d x \text {, where } x \geq 0$

$\text { Let } I=\int \frac{\sqrt{x}}{(1+x)^2} d x$

On putting, $x=\tan ^2 \theta \Rightarrow d x=2 \tan \theta \sec ^2 \theta d \theta$

$\begin{aligned} \therefore \quad & =\int \frac{(\tan \theta)\left(2 \tan \theta \sec ^2 \theta\right)}{\left(1+\tan ^2 \theta\right)^2} d \theta \\ & =\int \frac{2 \tan ^2 \theta \sec ^2 \theta}{\sec ^4 \theta} d \theta \\ & =2 \int \sin ^2 \theta d \theta \\ & =2 \int \frac{1-\cos 2 \theta}{2} d \theta \\ & =\theta-\frac{\sin 2 \theta}{2}+C \\ & =\theta-\frac{\tan \theta}{1+\tan ^2 \theta}+C=\tan ^{-1} \sqrt{x}-\frac{\sqrt{x}}{1+x}+C \end{aligned} $

$ \begin{aligned} & \therefore f(3)-f(1)=\left(\tan ^{-1} \sqrt{3}-\frac{\sqrt{3}}{1+3}+C\right) \\ & -\left(\tan ^{-1} \sqrt{1}-\frac{\sqrt{1}}{1+1}+C\right) \\ & =\frac{\pi}{3}-\frac{\pi}{4}+\frac{1}{2}-\frac{\sqrt{3}}{4}=\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4} \end{aligned} $

$I = \int {{1 \over {{{\left( {{{x - 1} \over {x + 2}}} \right)}^{3/4}}{{(x + 2)}^2}}}dx} $

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

$ \Rightarrow I = {1 \over 3}\int {{1 \over {{t^{3/4}}}}dt = {1 \over 3}\left( {{{{t^{1/4}}} \over {1/4}}} \right) + C} $

$ = {{{4^{{t^{1/4}}}}} \over 3} + C = {4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{1/4}} + C$

$f(x) = \int {{{{x^2}dx} \over {(1 + {x^2})(1 + \sqrt {1 + {x^2}} )}}} $

Let $x = \tan \theta $

$ \Rightarrow dx = {\sec ^2}\theta d\theta = (1 + {x^2})d\theta $

$f(x) = \int {{{{x^2}dx} \over {(1 + {x^2})(1 + \sqrt {1 + {x^2}} )}}} $

$ = \int {{{{{\tan }^2}\theta {{\sec }^2}\theta d\theta } \over {{{\sec }^2}\theta (1 + \sec \theta )}} = \int {{{{{\tan }^2}\theta d\theta } \over {1 + \sec \theta }}} } $

$ = \int {{{{{\sin }^2}\theta d\theta } \over {\cos \theta (1 + \cos \theta )}}} $

$ = \int {{{1 - {{\cos }^2}\theta d\theta } \over {\cos \theta (1 + \cos \theta )}}} $

$ = \int {{{(1 - \cos \theta )d\theta } \over {\cos \theta }}} $

$ = \int {\sec \theta d\theta - \int {d\theta } } $

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

$\therefore$ $f(0) = \log (0 + \sqrt {1 + 0} ) - {\tan ^{ - 1}}(0) + C$

$0 = \log 1 - 0 + C$

$ \Rightarrow C = 0$

$\therefore$ $f(1) = \log (1 + \sqrt {1 + {1^2}} ) - {\tan ^{ - 1}}(1)$

$ = \log (1 + \sqrt 2 ) - {\pi \over 4}$

Let $I = \int {{1 \over {1 - 2\sin x}}dx} $

On putting $\sin x = {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}$, we get

$I = \int {{1 \over {1 - {{4\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}}}dx = \int {{{1 + {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2} - 4\tan {x \over 2}}}dx} } $

$ = \int {{{{{\sec }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2} - 4\tan {x \over 2}}}dx} $

$ = 2\int {{1 \over {{{\left( {\tan {x \over 2} - 2} \right)}^2} - {{(\sqrt 3 )}^2}}}d\left( {\tan {x \over 2}} \right)} $

$ = 2\int {{{dt} \over {{{(t - 2)}^2} - {{(\sqrt 3 )}^2}}}} $, where $t = \tan {x \over 2}$

$ = 2\,.\,{1 \over {2\sqrt 3 }}\log \left| {{{t - 2 - \sqrt 3 } \over {t - 2 + \sqrt 3 }}} \right| + c = {1 \over {\sqrt 3 }}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c$

We have,

$I = \int {{{8{x^{43}} + 13{x^{38}}} \over {{{({x^{13}} + {x^5} + 1)}^4}}}dx} $

$ = \int {{{8{x^{43}} + 13{x^{38}}} \over {{x^{52}}\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}}dx} $

$ = \int {{{8\left( {{1 \over {{x^9}}}} \right) + 13\left( {{1 \over {{x^{14}}}}} \right)} \over {{{\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}^4}}}dx} $

Put, $1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}} = t$

$ \Rightarrow \left( {{{ - 8} \over {{x^9}}} - {{13} \over {{x^{14}}}}} \right)dx = dt$

$\therefore$ $I = - \int {{{dt} \over {{t^4}}} = {1 \over {3{t^3}}} + C} $

$\eqalign{ & \Rightarrow I = {1 \over {3{{\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}^3}}} + C = {{{x^{39}}} \over {3{{({x^{13}} + {x^5} + 1)}^3}}} + C \cr & \cr} $