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).
 A: 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 $$\begin{align}\int \frac{u^2+1}{2u^{101}}\,du&=\frac{1}{2}\int u^{-99}+u^{-101}\,du\\&=\frac{1}{2}\left(\frac{u^{-98}}{-98}-\frac{u^{-100}}{100}\right)+C\\&=-\frac{1}{196(x+\sqrt{x^2+1})^{98}}-\frac{1}{200(x+\sqrt{x^2+1})^{100}}+C\end{align}$$
A: $\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}$
Substitute $x = \sinh(u)$
$\int \frac{\cosh(u)du}{(\sinh(u) +\sqrt{\sinh^2(u)+1})^{99}} = $
$\int \frac{e^{u} + e^{-u}}{2}e^{-99u} = $
$\frac{1}{2}\int e^{-98u} + e^{-100u} = $
$\frac{-1}{196}e^{-98u} + \frac{-1}{200}e^{-100u} + C$
Where $u = \sinh^{-1}(x)$.
A: Let's start with $x=\sinh(t)$ then :
$$\int \frac{\cosh(t)dt}{(\sinh(t)+\cosh(t))^{99}}=\int \frac{\cosh(t)dt}{ e^{99t}}=\frac 12\int e^{-98t}+e^{-100t}\,dt$$
$$=-\frac 12\left(\frac {e^{-98t}}{98}+\frac{e^{-100t}}{100}\right)$$
so that :
$$\int \frac{dx}{\left(x+\sqrt{x^2+1}\right)^{99}}=-\frac 12\left(\frac {e^{-98\cdot\mathrm{asinh}(x)}}{98}+\frac{e^{-100\cdot\mathrm{asinh}(x)}}{100}\right)+C$$
that we may 'simplify' as Alpha.
We may too use $\mathrm{asinh}(x)=\log(x+\sqrt{x^2+1})$ to rewrite this simply as :
$$\int \frac{dx}{\left(x+\sqrt{x^2+1}\right)^{99}}=-\frac 12\left(\frac {\left(x+\sqrt{x^2+1}\right)^{-98}}{98}+\frac{\left(x+\sqrt{x^2+1}\right)^{-100}}{100}\right)+C$$
or
$$\int \frac{dx}{\left(x+\sqrt{x^2+1}\right)^{99}}=-\frac 1{2\left(x+\sqrt{x^2+1}\right)^{100}}\left(\frac {\left(x+\sqrt{x^2+1}\right)^2}{98}+\frac 1{100}\right)+C$$
A: 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 $\sqrt{x^2+1}$.
