Solving $\int_{0}^{\infty} \frac{x^2}{x^8+5}dx$? $$\int_{0}^{\infty} \frac{x^2}{x^8+5}\  \mathrm{d}x?$$
It's very hard for me to solve above integral. Please help me. 
 A: All integrals of the form $\displaystyle\int_0^\infty\dfrac{x^{k-1}}{x^n+a^n}~dx$ can be shown to equal $a^{k-n}~\dfrac\pi n~\csc\bigg(k~\dfrac\pi n\bigg)$ using the following method: First, substitute $x=at,$ then let $u=\dfrac1{t^n+1},$ and recognize the expression of the beta function in the new integral. Lastly, employ Euler's reflection formula for the $\Gamma$ function to establish the final result.
A: If one knows the Euler $\Gamma$ function,
$$
\Gamma(s)=\int_0^\infty u^{s-1} e^{-u}\:{\rm{d}}u, \quad s>0,
$$
and some of its properties then one may write
$$
\begin{align}
&\int_{-\infty}^{+\infty} \frac{x^2}{x^8+5} \:{\rm{d}}x
\\\\&=2\int_0^{+\infty} \frac{x^2}{x^8+5} \:{\rm{d}}x
\\\\&=2\int_0^\infty\left(\int_0^{+\infty} x^2e^{-(x^8+5)t} \:{\rm{d}}t\right){\rm{d}}x
\\\\&=2\int_0^\infty e^{-5t}\left(\int_0^\infty x^2e^{-x^8 t}\:{\rm{d}}x\right)\:{\rm{d}}t
\\\\&=\frac14\int_0^\infty t^{-3/8}e^{-5t}\left(\int_0^\infty u^{3/8-1} e^{-u}{\rm{d}}u\right)\:{\rm{d}}t\quad (x=u^{-1/8}t^{-1/8})
\\\\&=\frac1{4\cdot5^{5/8}}\left(\int_0^\infty v^{-3/8}e^{-v}\:dv\right)\left(\int_0^\infty u^{3/8-1} e^{-u}{\rm{d}}u\right)\quad (v=5t)
\\\\&=\frac1{4\cdot5^{5/8}} \Gamma\left(1-\frac38\right)\Gamma\left(\frac38\right)
\\\\&=\frac{\pi}{4\cdot5^{5/8}\sin \frac{3\pi}8}
\\\\&=\frac{\pi}{2\cdot5^{5/8}\sqrt{2+\sqrt{2}}}.
\end{align}
$$
A: We have:
$$ I=\int_{-\infty}^{+\infty}\frac{x^2}{x^8+5}\,dx = 2\int_{0}^{+\infty}\frac{x^2}{x^8+5}\,dx = \frac{2}{3}\int_{0}^{+\infty}\frac{dz}{z^{8/3}+5}{\phantom{deleted excess dz}} $$
by setting $x=z^{1/3}$ in the last step. Now, by setting $z=5^{3/8}w$,
$$ I = \frac{2}{3\cdot 5^{5/8}}\int_{0}^{+\infty}\frac{dw}{w^{8/3}+1}, $$
but for any $\alpha > 1$ we have:
$$ \int_{0}^{+\infty}\frac{dw}{w^\alpha+1} = \frac{\pi}{\alpha \sin\left(\frac{\pi}{\alpha}\right)} $$
due to Euler's beta function and the $\Gamma$ reflection formula. It follows that:

$$ \int_{-\infty}^{+\infty}\frac{x^2}{x^8+5}\,dx = \color{red}{\frac{\pi}{2^{5/4}\cdot 5^{5/8}\cdot\sqrt{1+\sqrt{2}}}}. $$

