Computing complex integrals 2 How do we compute the following kind of integrals, I think there are couple of ways but just one is enough for me I can study the way and solve the rest.
$
\begin{array}{l}
{\displaystyle\int_{0}^{1}{\mathrm{d}x \over
\left[\,x^{2}\left(1 - x\right)\,\right]^{1/3}}}
\\[2mm]
{\displaystyle\int_{0}^{1}{\mathrm{d}x \over
\left(x + 1\right)\left[\,x^{2}\left(1 - x\right)\,\right]^{1/3}}}
\end{array}
$
There are similar more questions like this but I want to solve them by my own. I would appreciate any help, thanks in advance.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[#ffd,10px]{\ds{%
\int_{0}^{1}{\dd x \over \pars{1 + x}\bracks{x^{2}\pars{1 - x}}^{1/3}}}}
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\int_{\infty}^{1}{-\dd x/x^{2} \over
\pars{1 + 1/x}\bracks{x^{-2}\pars{1 - 1/x}}^{1/3}}
\\[5mm] = &\
\int_{1}^{\infty}{\dd x \over \pars{1 + x}\pars{x - 1}^{1/3}}
\,\,\,\stackrel{x - 1\ \mapsto\ x}{=}\,\,\,
\int_{0}^{\infty}{x^{-1/3} \over 2 + x}\,\dd x
\,\,\,\stackrel{x/2\ \mapsto\ x}{=}\,\,\,
2^{-1/3}\int_{0}^{\infty}{x^{-1/3} \over 1 + x}\,\dd x
\\[5mm] \stackrel{x\ =\ 1/t - 1}{=}\,\,\,&
2^{-1/3}\int_{1}^{0}\pars{{1 \over t} - 1}^{-1/3}\,t
\pars{-\,{\dd t \over t^{2}}}
\\[5mm] = & 2^{-1/3}\
\overbrace{\int_{0}^{1}t^{-2/3}\pars{1 - t}^{-1/3}\,\dd t}
^{\ds{\mrm{B}\pars{{1 \over 3},{2 \over 3}}}}\qquad
\pars{~\mrm{B}:\ Beta\ Function~}
\\[5mm] = &\
2^{-1/3}\,{\Gamma\pars{1/3}\Gamma\pars{2/3} \over \Gamma\pars{1}}
\qquad\pars{~\Gamma:\ Gamma\ Function.\ \Gamma\pars{1} = 1~}
\\[5mm] = &\
2^{-1/3}\,{\pi \over \sin\pars{\pi/3}}
\qquad\pars{~Euler\ Reflection\ Formula~}
\\[5mm] = &\
\bbx{{2^{2/3} \over \root{3}}\,\pi} \approx 2.8792
\end{align}
A: The first integral can be easily done with beta-function:
$\mathcal{J}=\int_0^1 {dx\over (x^2 (1-x))^{1/3}}=\int_0^1 x^{-2/3}(1-x)^{-1/3}\,dx=\beta\left(\frac{1}{3},\frac{2}{3}\right)=\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)=\frac{\pi}{\sin\left(\frac{1}{3}\pi\right)}=\frac{2\pi}{\sqrt{3}}$
A: For the second integral, one may show that the segment cut between 0 and 1 on the real axis is sufficient to properly define the function $f(z)=(z+1)^{-1}\left( z^2(1-z)\right)^{-1/3}$. It has a pole at $z=-1$. Then as $zf(z)\to 0$ when $z\to \infty$, the integral
$$J=\int_\gamma f(z)\,dz$$ where $\gamma$ is the dog bone contour around the cut segment in the clockwise direction, can be expressed using the residue theorem as $$J=2i\pi\, \text{Res}\left(f(z),z=-1\right)$$ It can also be expressed as $$J=\int_0^1 f(x)\,dx+e^{4i\pi/3}\int_1^0f(z)\,dz=(1-e^{4i\pi/3})I$$
The residue is $2^{-1/3}e^{-2i\pi/3}$, then $$I=2i\pi\frac{2^{-1/3}e^{-2i\pi/3}}{(1-e^{4i\pi/3})}=\frac{2^{2/3}\pi}{\sqrt 3}$$
To be precise on the determinations for the phase used in the above calculation: $$\left( z^2\left( 1-z \right) \right)^{-1/3}=\left|z^2\left( 1-z \right)\right|^{-1/3}e^{-\frac{2\theta_0+\theta1}{3}}$$ where $\theta_0$ and $\theta_1$ correspond to the polar angles of $z$ and $1-z$. In the contour from above the segment to below, the integration point turned around $z=1$, so the phase angle of $1-z$ took $-2\pi$. The point $z=-1$ corresponds to $\theta_0=\pi$ and $\theta_1=0$.
