Question

Solve :
${\int }_{\infty }^0\frac{x}{(x+\sqrt{1+x^2}{)}^n}dx$; where $nϵN,n>2$

Answer 1

${\int }_{\infty }^1\frac{xdx}{x+\sqrt{1+x^2}}=I$

put $t=\sqrt{1+x^2}+x\Rightarrow \frac{1}{t}=\frac{1}{a+x^2+x}\times \frac{(\sqrt{1+x^2}-x)}{(\sqrt{1+x^2}-x)}=\frac{\sqrt{1+x^2}-x}{1+x^2-x^2}=\sqrt{1+x^2}-1$

$\therefore t+\frac{1}{t}=\sqrt{1+x^2}+x+\sqrt{1+x^2}-x=2\sqrt{1+x^2}$ ......... $\left(1\right)$

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

$dt(1+\frac{2x}{2\sqrt{1+x^2}})dx=\left(\frac{1+x}{\sqrt{1+x^2}}\right)dx=(1+\frac{\frac{\left(\right(t-\frac{1}{t})}{2}}{\frac{(t+\frac{1}{t})}{2}})$ (from $\left(1\right)$ & $\left(2\right)$)

$dt=[1+\left(\frac{t^2-+1}{t^2+1}\right)]dx=\frac{2t^2}{(t^2+1)}dx$

$\Rightarrow dx=\left(\frac{t^2+1}{2t^2}\right)dt$ if$x=0,t=1$

if $x=0,t=0$

$I={\int }_{\infty }^1\frac{(t-\frac{1}{t})}{2t^n}\times \frac{(t^2+1)}{2t^2}dt$

$I={\int }_{\infty }^1\frac{(t^2-1)(t^2+1)dt}{4t^{n+3}}$

$=\frac{1}{4}{\int }_{\infty }^1\frac{(t^n-1)}{t^{n+3}}dt$

$=\frac{1}{4}{\int }_{\infty }^1\frac{t^4}{t^{n+3}}dt-\frac{1}{4}{\int }_{\infty }^1\frac{1}{t^{n+3}}dt$

$=\frac{1}{4}{\int }_{\infty }^1{t}^{n+1}dt-\frac{1}{4}{\int }_{\infty }^1{t}^{n-3}dt$

${\Rightarrow \frac{1}{4}\frac{t^{n+2}}{(-n+2)}]}_{\infty }^1-\frac{1}{4}{\frac{t^{n-3+1}}{(n-3+1)}]}_{\infty }^1$

$I\Rightarrow \frac{1}{4}[\frac{1}{2-n}+\frac{1}{2+n}]\Rightarrow \frac{1}{4}\left[\frac{2+n+2-n}{4-n^2}\right]\Rightarrow \frac{1}{4-n^2}$

$\therefore I=\frac{1}{4-n^2}$