Show that $\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n}$ converges and is equal to $\frac{n}{n^2-1}$ Testing convergence of  $$I=\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n},$$ show that $$\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n}=\frac{n}{n^2-1}$$
I thought to put $$x=\tan{z}, \text{ then } \quad I=\int_0^{\pi/2}\frac{\sec^2zdz}{(\tan{z}+\sec{z})^n}$$
It is difficult to investigate convergence. Help is earnestly solicited. 
 A: As for the convergence, the integrand is continuous on $[0,\infty)$ and thus only the behavior as $x\to\infty$ matters. But since
$$ \frac{1}{(x + \sqrt{1+x^2})^n} \sim \frac{1}{(2x)^n} \qquad \text{as} \quad  x\to\infty, $$
by the (limit) comparison test, the integran converges exactly when $n > 1$.
In order to compute the exact value, it would be better to work with the substitution $x = \sinh t$. Then $x+\sqrt{1+x^2} = e^t$ and hence
$$ I = \int_{0}^{\infty} \cosh(t) e^{-nt} \, dt = \int_{0}^{\infty} \frac{e^{-(n-1)t} + e^{-(n+1)t}}{2} \, dt. $$
Can you proceed from here?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{x \equiv {1 - t^{2} \over 2t}}$:

$$
\bbx{\begin{array}{l}
\mbox{The integrand is}\ \ds{\stackrel{\mrm{as}\ x\ \to\ \infty}{\sim}
{x^{-n} \over 2^{n}}\ \mbox{such that the integral converges whenever}\
\color{red}{n > 1}}
\end{array}}
$$
Its evaluation is straightforward. Namely,
\begin{align}
\left.\int_{0}^{\infty}{\dd x \over \pars{x + \root{x^{2} + 1}}^{n}}
\right\vert_{\ n\ >\ 1} & =
{1 \over 2}\int_{0}^{1}\pars{t^{n - 2} + t^{n}}\dd t =
{1 \over 2}\pars{{t^{n - 1} \over n - 1} + {t^{n + 1} \over n + 1}}_{0}^{1}
\\[5mm] & =
{1 \over 2}\pars{{1 \over n - 1} + {1 \over n + 1}} = \bbx{n \over n^{2} - 1}
\end{align}
A: Using twice the fact that $\begin{aligned} & \frac{d}{d x} \frac{1}{(\tan \theta+\sec \theta)^n} = -\frac{n \sec \theta}{(\tan \theta+\sec \theta)^n}\end{aligned} $ via integration by parts, we have a beautiful recursive relation for $I$ as below:
$$
\begin{aligned}
I& = \int_0^{\infty} \frac{d x}{\left(x+\sqrt{1+x^2}\right)^n}\\& \stackrel{x=\tan \theta }{=} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 \theta d \theta}{(\tan \theta+\sec \theta)^n} d x . \\&=-\frac{1}{n}\int_0^{\frac{\pi}{2}} \sec \theta \, d\left(\frac{1}{(\tan \theta+\sec \theta)^n}\right) \\
& =\left[-\frac{\sec \theta}{n(\tan \theta+\sec \theta)^n}\right]_0^{\frac{\pi}{2}}+\frac{1}{n} \int_0^{\frac{\pi}{2}} \frac{\sec \theta\tan \theta}{(\tan \theta+\sec \theta)^n} d \theta \\
& =\frac{1}{n}-\frac{1}{n^2} \int_0^{\frac{\pi}{2}} \tan \theta d\left(\frac{1}{(\tan \theta+\sec \theta)^n}\right) \\
& =\frac{1}{n}-\left[\frac{1}{n^2} \frac{\tan \theta}{(\tan \theta+\sec \theta)^n}\right]_0^{\frac{\pi}{2}}+\frac{1}{n^2}  \int_d^{\frac{\pi}{2}} \frac{\sec ^2 \theta}{(\tan \theta+\sec \theta)^n} d \theta \\
& =\frac{1}{n}+\frac{1}{n^2}I
\end{aligned}
$$
Now we can conclude that
$$\boxed{I_n=\frac{n}{n^2-1}}$$
