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$$ \mbox{If}\ n>1,\ \mbox{then prove that}\quad\int_{0}^{\infty}{\mathrm{d}x \over \left(x + \,\sqrt{\, x^{2} + 1\,}\,\right)^{n}} = {n \over n^{2} - 1} $$

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.

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Could someone give me little hint so that I could proceed in this question.

Hint. One may perform the change of variable $$ x \in [0,\infty),\quad x=\sinh u \implies x+\sqrt{x^2+1}=e^u, \quad dx=\cosh u\:du, $$ giving $$ \int^{\infty}_{0} \frac{dx}{(x+\sqrt{x^2+1})^n}=\int^{\infty}_{0} e^{-nu}\cosh u\:du. $$ Can you take it from here?

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    $\begingroup$ [+1] Supplementary hint : If you know Laplace Transform, you have an immediate result... $\endgroup$ – Jean Marie Dec 21 '16 at 20:43
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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

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

$$=\frac{1}{2}\int_0^\infty\left[\exp(-(n-1)\theta+\exp(-(1+n)\theta)\right]d\theta=\frac{1}{2}\left[\frac{1}{n-1}+\frac{1}{n+1}\right]=\frac{n}{n^2-1}$$

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You asked for a hint, so here is one:

Let $$ u=x+\sqrt{1+x^2}. $$

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  • $\begingroup$ Why do someone want to delete my answer? The hint is really serious. It transforms the integral into the easiest possible. Please explain yourself! $\endgroup$ – mickep Dec 30 '16 at 18:14

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