Integrating $\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}$ I am bugged by this problem: how do I evaluate this?
$$\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}.$$
A closed form will be convenient and fine. Thanks (it does not seem particularly inpiring).

If you sub $u=x+\sqrt{x^2+1}$, that alone will make the integral a simple rational integral. We can solve for $x$ in terms of $u$:
$$\begin{align}u - x & =\sqrt{x^2+1}\\
u^2 - 2ux + x^2& =x^2+1\\
- 2ux& =1-u^2\\
x & = \frac{u^2-1}{2u}\\
dx & = \frac{2u(2u)-2(u^2-1)}{4u^2} du\\
dx & = \frac{u^2+1}{2u^2} du
\end{align}$$
So you have $$\begin{align}\int \frac{u^2+1}{2u^{101}}\,du&=\frac{1}{2}\int u^{-99}+u^{-101}\,du\\&=\frac{1}{2}\left(\frac{u^{-98}}{-98}-\frac{u^{-100}}{100}\right)+C\\&=-\frac{1}{196(x+\sqrt{x^2+1})^{98}}-\frac{1}{200(x+\sqrt{x^2+1})^{100}}+C\end{align}$$
