Prove $\int_0^1 \frac{dx}{(x-2) \sqrt[5]{x^2{(1-x)}^3}} = -\frac{2^{11/10} \pi}{\sqrt{5+\sqrt{5}}}$ $$
\mbox{Prove}\quad
\int_{0}^{1}{\mathrm{d}x \over
\left(\,{x - 2}\,\right)\,
\sqrt[\Large 5]{\,x^{2}\,\left(\,{1 - x}\,\right)^{3}\,}\,}
=
-\,{2^{11/10}\,\pi \over \,\sqrt{\,{5 + \,\sqrt{\,{5}\,}}\,}\,}
$$

*

*Being honest I havent got a clue where to start.  I dont think any obvious substitutions will help ($x \to 1-x, \frac{1}{x}, \sqrt{x},$ more).

*The indefinite integral involves hypergeometric function so some miracle substitution has to work with the bounds I suspect.

*Maybe gamma function is involved some how ??.

If anyone has an idea and can provide help I would appreciate it.

$$
\begin{aligned}
\text { Let } x&=\sin ^2 \theta, \quad \textrm{ then } d x=2 \sin \theta \cos \theta d \theta, \textrm{ and }\\
I&=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\left(\sin ^2 \theta-2\right) \sqrt[5]{\sin ^4 \theta \cos ^6 \theta}}\\
&=2 \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{1}{5}} \theta  d \theta}{\cos ^2 \theta\left(\tan ^2 \theta-2 \sec ^2 \theta\right)}\\
&=-2 \int_0^{\infty} \frac{t^{\frac{1}{5}}}{t^2+2} d t ,\quad \textrm{ where }t=\tan \theta\\
&=-2\left(\frac{2^{\frac{1}{10}} \pi}{\sqrt{5+\sqrt{5}}}\right) \quad \textrm{(via beta function)} \\
&=-\frac{2^{\frac{11}{10}} \pi}{\sqrt{5+\sqrt{5}}}
\end{aligned}
$$
