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.

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?