# 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).

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Try $x = \sinh(u)$? – Bitrex Aug 19 '12 at 7:16
Would it help to multiply top and bottom by $(\sqrt{x^2+1}-x)^{99}$? – Mike Aug 19 '12 at 7:26

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}

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$\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}$

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 $u = \sinh^{-1}(x)$.

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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 $$\int \frac{dx}{\left(x+\sqrt{x^2+1}\right)^{99}}=-\frac 1{2\left(x+\sqrt{x^2+1}\right)^{100}}\left(\frac {\left(x+\sqrt{x^2+1}\right)^2}{98}+\frac 1{100}\right)+C$$

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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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