An integral. Nothing too fancy... or not?

**Pranas** · September 13th 2011, 12:21 PM

Hi everyone.
Sorry for such a lame thread, but I feel extremely rusty with my math...

$\ell = 2 \cdot \int\limits_0^{ + \infty } {\frac{{\sqrt {{x^2} + 1} }}{{{{(x + 1)}^3}}}} dx$

Can you give a hint how to make this one nice and easy?

P.S. I don't think it's important, but as a matter of fact it arose from evaluating the length of a curve $\sqrt x + \sqrt y = 1$

Thanks!

**Ackbeet** · September 13th 2011, 12:47 PM

I don't think that integral is easy. Mathematica yields

$2\int_{0}^{\infty}\frac{\sqrt{x^{2}+1}}{(x+1)^{3}} \,dx=-\frac{\text{MeijerG}[\{\{\frac{1}{2},1,2\},\{\}\},\{\{\frac{1}{2},\frac {3}{2},2\},\{\}\},1]}{\pi^{3/2}},$

where

$\text{MeijerG}[\{\{a_{1},\dots,a_{n}\},\{a_{n+1},\dots,a_{p}\}\}, \{\{b_{1},\dots,b_{m}\},\{b_{m+1},\dots,b_{q}\}\}, z]$ is the MeijerG function

$G_{p\;q}^{m\,n}}\left(z\Bigg|\begin{matrix}a_{1}, \dots,a_{p}\\ a_{1},\dots,b_{q}\end{matrix}\right).$

Not very pretty, if you ask me.

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