<s>[INST] 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).
 [/INST] I like Bitrex's answer best, and my other answer next, but here is one without any substitution. Multiply by $\frac{(\sqrt{x^2+1}-x)^{99}}{(\sqrt{x^2+1}-x)^{99}}$ and you have $$\int\frac{(\sqrt{x^2+1}-x)^{99}}{1}\,dx=\int(\sqrt{x^2+1}-x)^{99}\,dx$$ The next part isn't fun to write out, but you could use the binomial theorem to expand the 99th power. $$\int\sum_{k=0}^{99}\binom{99}{k}(-x)^{99-k}(\sqrt{x^2+1})^k\,dx$$ For each $k$, you have a term that is easy to antidifferentiate. 
$$\begin{align}
&\int\sum_{k=0}^{49}\binom{99}{2k}(-x)^{99-2k}(\sqrt{x^2+1})^{2k}\,dx+\int\sum_{k=0}^{49}\binom{99}{2k+1}(-x)^{99-2k-1}(\sqrt{x^2+1})^{2k+1}\,dx\\
=&-\int\sum_{k=0}^{49}\binom{99}{2k}x^{99-2k}(x^2+1)^{k}\,dx+\int\sum_{k=0}^{49}\binom{99}{2k+1}x^{98-2k}(x^2+1)^{k}\sqrt{x^2+1}\,dx\\
=&-\int\sum_{k=0}^{49}\binom{99}{2k}x^{99-2k}\sum_{j=0}^kx^{2j}\,dx+\int\sum_{k=0}^{49}\binom{99}{2k+1}x^{98-2k}\sum_{j=0}^kx^{2j}\sqrt{x^2+1}\,dx\\
=&-\sum_{k=0}^{49}\binom{99}{2k}\sum_{j=0}^k\frac{x^{2j+100-2k}}{2j+100-2k}+\sum_{k=0}^{49}\binom{99}{2k+1}\sum_{j=0}^k\int x^{2j+98-2k}\sqrt{x^2+1}\,dx
\end{align}$$
The last integral can be computed using a reduction formula for $x^n\sqrt{x^2+1}$ (effectively, integration by parts with $u=x^{n-1}$ and $dv=x\sqrt{x^2-1}\,dx$) and the antiderivative for $\sqrt{x^2+1}$.
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