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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over \pars{x - 2}\root[\Large 5]{x^{2}\pars{1 - x}^{3}}}= -\,{2^{11/10}\,\pi \over \root{5 + \root{5}}}}:\ {\Large ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over \pars{x - 2}\root[\Large 5]{x^{2}\pars{1 - x}^{3}}}}
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\int_{\infty}^{1}{-\,\dd x/x^{2} \over \pars{1/x - 2}\pars{1/x}^{2/5}
\pars{1 - 1/x}^{3/5}}
\\[5mm] = &\
\int_{1}^{\infty}{\pars{x - 1}^{-3/5} \over 1 - 2x}\,\dd x
\,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\,
-\,{1 \over 2}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x
\end{align}

Lets consider $\ds{\oint_{C}{z^{-3/5} \over z - 1/2}\,\dd z}$ where $\ds{C}$ is a key-hole contour which "takes care" of the principal branch of $\ds{\rule{0cm}{8mm}z^{-3/5}}$ ( with a branch-cut along
$\ds{\left(-\infty,0\right]}$ ). Namely,

\begin{align}
2\pi\ic\bracks{\pars{1 \over 2}^{-3/5}} & =
\int_{-\infty}^{0}{\pars{-x}^{-3/5}\expo{-3\pi\ic/5}
 \over x - 1/2}\,\dd x + \int_{0}^{-\infty}{\pars{-x}^{-3/5}\expo{3\pi\ic/5}
 \over x - 1/2}\,\dd x
\\[5mm] & =
-\expo{-3\pi\ic/5}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x +
\expo{3\pi\ic/5}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x
\\[5mm] & =
2\ic\sin\pars{3\pi \over 5}
\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x
\\[5mm] \implies & 
\bbox[10px,#ffd]{-\,{1 \over 2}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x =
- 2^{3/5}\pi\csc\pars{3\pi \over 5}} \approx -2.5034
\end{align}
A: $$
\begin{aligned}
\text { Let } x&=\sin ^2 \theta, \quad \textrm{ then } d x=2 \sin \theta \cos \theta d \theta, \textrm{ and }\\
I&=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\left(\sin ^2 \theta-2\right) \sqrt[5]{\sin ^4 \theta \cos ^6 \theta}}\\
&=2 \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{1}{5}} \theta  d \theta}{\cos ^2 \theta\left(\tan ^2 \theta-2 \sec ^2 \theta\right)}\\
&=-2 \int_0^{\infty} \frac{t^{\frac{1}{5}}}{t^2+2} d t ,\quad \textrm{ where }t=\tan \theta\\
&=-2\left(\frac{2^{\frac{1}{10}} \pi}{\sqrt{5+\sqrt{5}}}\right) \quad \textrm{(via beta function)} \\
&=-\frac{2^{\frac{11}{10}} \pi}{\sqrt{5+\sqrt{5}}}
\end{aligned}
$$
A: Hint:
Substitute $x \rightarrow\frac{1}{x-1}$. We'll get: $$-\int_0^\infty \dfrac{x^{-3/5}dx}{(2x+1)}$$
Can you continue from here?
