Evaluation of the integral $\int^{\infty}_{0} \frac{dx}{(x+\sqrt{x^2+1})^n}$ $$
\mbox{If}\ n>1,\ \mbox{then prove that}\quad\int_{0}^{\infty}{\mathrm{d}x \over
\left(x + \,\sqrt{\, x^{2} + 1\,}\,\right)^{n}} = {n \over n^{2} - 1}
$$
Could someone give me little hint so that I could proceed in this question. I tried putting $x = \tan\left(A\right)$ but it did not work out.

Claim:
$$\int^{\infty}_{0} \frac{dx}{(x+\sqrt{x^2+1})^n}=\frac{n}{n^2-1}$$
Proof:


*

*Let $x=\sinh\theta=\frac{1}{2}\left(\exp\theta-\exp(-\theta)\right)$. 

*Then 


*

*$dx=\frac{1}{2}\left(\exp\theta+\exp(-\theta)\right)d\theta$, and

*$(x+\sqrt{x^2+1})=\exp\theta$, so we have



$$\int^{\infty}_{0} \frac{dx}{(x+\sqrt{x^2+1})^n}=\int^{\infty}_{0} \frac{1}{2}\left(\exp\theta+\exp(-\theta)\right)\cdot \exp(-n\theta)d\theta$$
$$=\frac{1}{2}\int_0^\infty\left[\exp(-(n-1)\theta+\exp(-(1+n)\theta)\right]d\theta=\frac{1}{2}\left[\frac{1}{n-1}+\frac{1}{n+1}\right]=\frac{n}{n^2-1}$$
