Integral of $\frac{x}{\sqrt{1+x^5}}$ I am trying to calculate the following integral:
$\displaystyle\int_0^\infty \frac{x}{\sqrt{1+x^5}}\,  dx$
But I can't seem to find a primitive for that function. I was trying to find a good substitution, but was unable to. Also, attempting to use parts becomes a dead end. What can I do?
 A: $$\int_{0}^{+\infty}\frac{x}{\sqrt{1+x^5}}\,dx = \frac{2}{5}\int_{0}^{+\infty}x^{-1/5}(1+x^2)^{-1/2}\,dx = \frac{2}{5}\int_{0}^{\pi/2}\sin\theta^{-1/5}\cos\theta^{-4/5}\,d\theta,$$
$$\int_{0}^{+\infty}\frac{x}{\sqrt{1+x^5}}\,dx=\frac{2}{5}\int_{0}^{1}t^{-1/5}(1-t^2)^{-9/10}\,dt=\frac{1}{5}\int_{0}^{1}u^{-3/5}(1-u)^{-9/10}\,du,$$
$$\int_{0}^{+\infty}\frac{x}{\sqrt{1+x^5}}\,dx=\frac{1}{5}\frac{\Gamma(2/5)\,\Gamma(1/10)}{\Gamma(1/2)}=\frac{\Gamma(2/5)\,\Gamma(1/10)}{5\sqrt{\pi}}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\
\overbrace{\int_{0}^{\infty}{x \over \root{1 + x^{5}}}
\,\dd x}^{\ds{\color{#c00000}{x^{5}\ \mapsto x}}}}
=\int_{0}^{\infty}{x^{1/5} \over \root{1 + x}}\,{1 \over 5}\,x^{-4/5}\,\dd x
={1 \over 5}\int_{0}^{\infty}{x^{-3/5} \over \root{1 + x}}\,\dd x
\\[5mm]&={1 \over 5}\ \overbrace{%
\int_{1}^{\infty}\pars{x - 1}^{-3/5}x^{-1/2}\,\dd x}
^{\ds{\color{#c00000}{x\ \mapsto\ {1 \over x}}}}
={1 \over 5}\int_{1}^{0}\pars{{1 \over x} - 1}^{-3/5}x^{1/2}\,
\pars{-\,{\dd x \over x^{2}}}
\\[5mm]&={1 \over 5}\int_{0}^{1}\pars{1 - x}^{-3/5}x^{-9/10}\,\dd x
={1 \over 5}\,{\Gamma\pars{2/5}\Gamma\pars{1/10} \over \Gamma\pars{1/2}}
=\color{#66f}{\large{\Gamma\pars{2/5}\Gamma\pars{1/10} \over 5\root{\pi}}}
\approx {\tt 2.3812}
\end{align}
A: You can make the substituion $x^{5}=u\Rightarrow dx=\frac{du}{5u^{4/5}}$ and obtain $$\frac{1}{5}\int_{0}^{\infty}\frac{u^{-3/5}}{\sqrt{1+u}}du=\frac{1}{5}\int_{0}^{\infty}\frac{u^{(2/5)-1}}{\left(1+u\right)^{(2/5)+(1/10)}}du=\frac{1}{5}\textrm{B}\left(\frac{2}{5},\frac{1}{10}\right)$$where B is the Beta function.
