Integrating, branch points I'm trying to learn branch cuts and its about $7$ days I'm thinking about doing one of Arfken problems:
Show below relation using the contour:

$$\int_0 ^1 \frac{dx}{(x^2 - x^3)^{\frac{1}{3}}} = \frac{2 \pi}{ 3^{\frac{1}{2}}}$$
I could calculate the integral on the big circle. And I know that because there is no pole in the region, I can put all integrals equal to zero. But I don't know how to calculate other integrations. I really need your help Mathematicians. Any answers are highly appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}} =
{2 \pi \over 3^{1/2}}:\ {\large ?}}$.

I'll consider the integral
$\ds{\oint_{\mc{DB}}z^{-2/3}\pars{1 - z}^{-1/3}\,\dd z}$. The contour $\mc{DB}$ is the dog-bone one of the OP picture.
$\ds{z^{-2/3}}$ and $\ds{\pars{1 - z}^{-1/3}}$ are given by
$$
\left\{\begin{array}{l}
\ds{z^{-2/3} = \verts{z}^{\,-2/3}\exp\pars{-\,{2 \over 3}\arg\pars{z}\ic}\,,
\qquad -\pi < \arg\pars{z} < \pi\,,\quad z \not= 0}
\\[2mm]
\ds{\pars{1 - z}^{-1/3} = \verts{1 - z}^{\,-1/3}\exp\pars{-\,{1 \over 3}\arg\pars{1 - z}\ic}\,,
\qquad 0 < \arg\pars{1 - z} < 2\pi\,,\quad z \not= 1}
\end{array}\right.
$$
By multiplying $\ds{z^{-2/3}}$ and $\ds{\pars{1- z}^{-1/3}}$, as given above,
we'll see that the product branch-cut is set along $\ds{\bracks{0,1}}$.

\begin{align}
\oint_{\mc{DB}}z^{-2/3}\pars{1 - z}^{-1/3}\,\dd z & =
\int_{1}^{0}x^{-2/3}\pars{1 - x}^{-1/3}\expo{-2\pi\ic/3}\,\dd x +
\int_{0}^{1}x^{-2/3}\pars{1 - x}^{-1/3}\,\dd x
\\[5mm] & =
2\ic\expo{-\pi\ic/3}\sin\pars{\pi \over 3}\int_{0}^{1}x^{-2/3}
\pars{1 - x}^{-1/3}\,\dd x
\\[5mm] & =
\ic\root{3}\expo{-\pi\ic/3}\int_{0}^{1}x^{-2/3}\pars{1 - x}^{-1/3}\,\dd x
\label{1}\tag{1}
\\[1cm]
\oint_{\mc{DB}}z^{-2/3}\pars{1 - z}^{-1/3}\,\dd z & =
-2\pi\ic\,\mrm{Res}_{\, z = 0}\pars{%
-\,{1 \over z^{2}}\,z^{2/3}\bracks{1 - {1 \over z}}^{-1/3}}
\\[5mm] & =
2\pi\ic\,\mrm{Res}_{\, z = 0}\pars{%
{1 \over z}\,\bracks{z - 1}^{-1/3}} = 2\pi\ic\,\verts{0 - 1}^{-1/3}
\expo{-\pi\ic/3}
\label{2}\tag{2}
\end{align}

\eqref{1} and \eqref{2} lead to
$$\bbx{\ds{%
\int_{0}^{1}x^{-2/3}\pars{1 - x}^{-1/3}\,\dd x = {2\pi \over 3^{1/2}}}}
$$


The whole procedure is explained in detail in a
  Methods of Contour Integration page.

A: The reason you are having trouble is that the picture is not all that great.  Better would be lines extending above and below the branch cut to the large circle.  In this case, the contour integral
$$\oint_C dz \, z^{-2/3} (z-1)^{-1/3} $$
is equal to, if the branch cut is the portion of the real axis $x \lt 1$:
$$\int_0^1 dx \, x^{-2/3} \, e^{-i \pi/3} (1-x)^{-1/3} + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \, \left (R e^{i \theta} \right )^{-2/3} \left (R e^{i \theta}-1 \right )^{-1/3} \\ + \int_1^0 dx \, x^{-2/3} \, e^{i \pi/3} (1-x)^{-1/3}$$
The factors $e^{\pm i \pi/3}$ come from the values of the phase $e^{\pm i \pi}$ above and below the branch cut.  This is applied when we change the sign of $z-1$ when $z \lt 1$.  Note that the contributions to the integral above and below the branch cut cancel when $z \ge 1$.
By Cauchy's theorem the contour integral is zero.  You can take the limit as $R \to \infty$ in the second integral (around the large circle) and find it becomes $i 2 \pi$ in this limit - this is the so-called residue at infinity.
The stated result follows.
