To Shafiqur Rahman. Integral exercise

Exercise:

Integrate, $\int{\frac{xdx}{1+\sqrt{x+3}+\sqrt{x+2}}}$

Solution: let ${{u}^{2}}=x+2\Leftrightarrow 2udu=dx\,\,\And \,\,{{u}^{2}}-2=x$

So \[\int{\frac{xdx}{1+\sqrt{x+3}+\sqrt{x+2}}=\int{\frac{2u\left( {{u}^{2}}-2 \right)}{1+\sqrt{{{u}^{2}}-2+3}+u}du}}=\int{\frac{2u\left( {{u}^{2}}-2 \right)}{1+u+\sqrt{{{u}^{2}}+1}}du}\]

But $\frac{2u\left( {{u}^{2}}-2 \right)}{1+u+\sqrt{{{u}^{2}}+1}}\times \frac{\left( 1+u \right)-\sqrt{{{u}^{2}}+1}}{\left( 1+u \right)-\sqrt{{{u}^{2}}+1}}=\frac{2u\left( {{u}^{2}}-2 \right)\left[ \left( 1+u \right)-\sqrt{{{u}^{2}}+1} \right]}{1+{{u}^{2}}+2u-{{u}^{2}}-1}$

$=\frac{2u\left( {{u}^{2}}-2 \right)\left[ \left( 1+u \right)-\sqrt{{{u}^{2}}+1} \right]}{2u}=\left( {{u}^{2}}-2 \right)\left[ \left( 1+u \right)-\sqrt{{{u}^{2}}+1} \right]$

$={{u}^{2}}+{{u}^{3}}-{{u}^{2}}\sqrt{{{u}^{2}}+1}-2-2u+2\sqrt{{{u}^{2}}+1}$

So $\int{\frac{2u\left( {{u}^{2}}-2 \right)}{1+u+\sqrt{{{u}^{2}}+1}}du=\frac{{{u}^{3}}}{3}+\frac{{{u}^{4}}}{4}-2u-{{u}^{2}}-\int{{{u}^{2}}\sqrt{{{u}^{2}}+1}\,du}+2\int{\sqrt{{{u}^{2}}+1}\,du}}$

Let $u=\tan \theta \Leftrightarrow du={{\sec }^{2}}\theta \,d\theta $

So $-\int{{{u}^{2}}\sqrt{{{u}^{2}}+1}\,du}=-\int{{{\tan }^{2}}\theta \sqrt{{{\tan }^{2}}\theta +1}{{\sec }^{2}}\theta \,d\theta =-\int{{{\tan }^{2}}\theta {{\sec }^{3}}\theta \,d\theta }}$

But ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$

So $-\int{{{\tan }^{2}}\theta {{\sec }^{3}}\theta \,d\theta }=-\int{\left( {{\sec }^{2}}\theta -1 \right){{\sec }^{3}}\theta }\,d\theta =-\int{{{\sec }^{5}}\theta \,d\theta }+\int{{{\sec }^{3}}\theta \,d\theta }$

Also $2\int{\sqrt{{{u}^{2}}+1}\,du}=2\int{\sqrt{{{\sec }^{2}}\theta }{{\sec }^{2}}\theta }d\theta =2\int{{{\sec }^{3}}\theta d\theta }$

So $\int{\frac{2u\left( {{u}^{2}}-2 \right)}{1+u+\sqrt{{{u}^{2}}+1}}du}=\frac{{{u}^{3}}}{3}+\frac{{{u}^{4}}}{4}-2u-{{u}^{2}}-\int{{{\sec }^{5}}\theta d\theta +3\int{{{\sec }^{3}}\theta d\theta }}$

By Reduction Formula for $\int{{}}{{\sec }^{5}}\theta d\theta $ we get :

$\int{{{\sec }^{5}}\theta }\,d\theta =\frac{{{\sec }^{4}}\theta \sin \theta }{4}+\frac{3}{4}\int{{{\sec }^{3}}\theta \,d\theta }$

Hence $\int{\frac{2u\left( {{u}^{2}}-2 \right)}{1+u+\sqrt{{{u}^{2}}+1}}du=\frac{{{u}^{3}}}{3}+\frac{{{u}^{4}}}{4}-2u-{{u}^{2}}-\frac{{{\sec }^{4}}\theta \sin \theta }{4}+\frac{9}{4}\int{{{\sec }^{3}}\theta d\theta }}$

Let’s start by parts integration for last term in our expression

$\int{{{\sec }^{3}}\theta d\theta }=\int{{{\sec }^{2}}\theta \sec \theta d\theta }$

Let $U=\sec \theta \,\And \,dV={{\sec }^{2}}\theta \Leftrightarrow dU=\sec \theta \tan \theta \,\,\And \,V=\tan \theta \,$

So $\int{{{\sec }^{3}}\theta d\theta =\sec \theta \tan \theta -\int{\tan \theta \sec \theta \tan \theta \,d\theta }}$

$=\sec \theta \tan \theta -\int{\left( {{\sec }^{2}}\theta -1 \right)\sec \theta \,d\theta }=\sec \theta \tan \theta -\int{{{\sec }^{3}}\theta d\theta +\int{\sec \theta d\theta }}$

Thus $\int{{{\sec }^{3}}\theta d\theta =\frac{\sec \theta \tan \theta }{2}+\frac{1}{2}\int{\sec \theta d\theta }}$

But $\int{\sec \theta \,d\theta =\int{\frac{\sec \theta \left( \tan \theta +\sec \theta  \right)}{\tan \theta +\sec \theta }d\theta =\int{\frac{d\left( \tan \theta +\sec \theta  \right)}{\tan \theta +\sec \theta }=\ln \left| \tan \theta +\sec \theta  \right|+c}}}$

Thus $\int{{{\sec }^{3}}\theta d\theta =\frac{\sec \theta \tan \theta }{2}+\frac{1}{2}\ln \left| \tan \theta +\sec \theta  \right|+c}$

So $\int{\frac{2u\left( {{u}^{2}}-2 \right)}{1+u+\sqrt{{{u}^{2}}+1}}du}=\frac{{{u}^{4}}}{4}+\frac{{{u}^{3}}}{3}-{{u}^{2}}-2u-\frac{{{\sec }^{4}}\theta \sin \theta }{4}+\frac{9}{8}\sec \theta \tan \theta +\frac{9}{8}\ln \left| \tan \theta +\sec \theta  \right|+c$

By Back word Substitution to get:

We take $u=\tan \theta \Rightarrow {{u}^{2}}={{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1\Rightarrow {{u}^{2}}+1={{\sec }^{2}}\theta \Rightarrow {{\left( {{u}^{2}}+1 \right)}^{2}}={{\sec }^{4}}\theta $

And $\sec \theta =\sqrt{{{u}^{2}}+1}$  also $u=\tan \theta =\frac{\sin \theta }{\cos \theta }=\sin \theta \sec \theta \Leftrightarrow \sin \theta =\frac{u}{\sec \theta }=\frac{u}{\sqrt{{{u}^{2}}+1}}$

So $\int{\frac{2u\left( {{u}^{2}}-2 \right)}{1+u+\sqrt{{{u}^{2}}+1}}du}=\frac{{{u}^{4}}}{4}+\frac{{{u}^{3}}}{3}-{{u}^{2}}-2u-\frac{1}{4}\frac{u{{\left( {{u}^{2}}+1 \right)}^{2}}}{\sqrt{{{u}^{2}}+1}}+\frac{9}{8}u\sqrt{{{u}^{2}}+1}+\frac{9}{8}\ln \left| u+\sqrt{{{u}^{2}}+1} \right|+c$

Also we have ${{u}^{2}}=x+2\Leftrightarrow u=\sqrt{x+2}\,\And \,{{u}^{4}}={{\left( x+2 \right)}^{2}},{{u}^{3}}=\left( x+2 \right)\sqrt{x+2}$

Therefore ,

$\int{\frac{xdx}{1+\sqrt{x+3}+\sqrt{x+2}}=\frac{{{\left( x+2 \right)}^{2}}}{4}+\frac{\sqrt{{{\left( x+2 \right)}^{3}}}}{3}-\left( x+2 \right)-2\sqrt{x+2}-\frac{1}{4}\frac{\sqrt{x+2}{{\left( x+3 \right)}^{2}}}{\sqrt{x+3}}}$

$+\frac{9}{8}\sqrt{x+2}\sqrt{x+3}+\frac{9}{8}\ln \left| \sqrt{x+2}+\sqrt{x+3} \right|+c$