<s>[INST] 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. 
 [/INST] 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
$$\boxed{I_n=\frac{n}{n^2-1}}$$
</s>