Complex analysis: contour integration Evaluate by contour integral:
$$\int_0^1{ dx\over (x^2-x^3)^\frac 13}$$
Should I go for some kind of substitution so that the limit changes to $0$ to $\pi/2$?
 A: Draw a dumbbell contour $C$ about the branch cut $z \in [0,1]$.  That is, two small circular segments about the branch points, and an upper and lower path connecting them above and below the real axis, respectively, as illustrated below:

We consider the integral
$$\oint_C dz \, f(z) = i 2 \pi \text{Res}_{z=0} \left [\frac{1}{z^2} f\left(\frac{1}{z}\right) \right ]$$
where
$$f(z) = z^{-2/3} (1-z)^{-1/3}$$
and the term on the right in the former equation is the residue at infinity.  This residue may be computed by seeing that
$$\frac{1}{z^2} f\left(\frac{1}{z}\right) = \frac{(z-1)^{-1/3}}{z}$$
so that
$$\text{Res}_{z=0} \left [\frac{1}{z^2} f\left(\frac{1}{z}\right) \right ] = (-1)^{-1/3}$$
I will elaborate on this in a bit.
Now, we define 
$$z^{-2/3} = e^{-(2/3) \log{z}}$$
such that $\arg{z} \in [-\pi,\pi)$.  This definition is a result of the original branch cut of this factor being $(-\infty,0]$.  Further define
$$(1-z)^{-1/3} = e^{-(1/3) \log{(1-z)}}$$
such that $\arg{(1-z)} \in [0,2\pi)$.  This definition is a result of the original branch cut of this factor being $[1,\infty)$. 
To summarize, on the lines above and below the real axis, $z=x \in [0,1]$ and therefore $\arg{z} = 0$.  On the line above the real axis, however, $\arg{(1-z)} = 2 \pi$.  Therefore above the real axis,  $z^{-2/3} (1-z)^{-1/3} =  x^{-2/3} (1-x)^{-1/3} e^{-i 2 \pi/3}$  Below the real axis,  $z^{-2/3} (1-z)^{-1/3} =  x^{-2/3} (1-x)^{-1/3}$ because there, $\arg{(1-z)} = 0$.  
Further, it should be clear that the integrals about the small circular arcs of radius $\epsilon$ around the branch points vanish as $\epsilon^{1/3}$.
Therefore, we may write
$$\left ( 1-e^{-i 2 \pi/3}\right) \int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi (-1)^{-1/3}$$
Because that residue was calculated from the $1-z$ term, then $-1=e^{i \pi}$ and we have
$$\int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi \frac{e^{-i \pi/3}}{1-e^{-i 2 \pi/3}} = \frac{\pi}{\sin{(\pi/3)}} = \frac{2 \pi}{\sqrt{3}}$$
A: Here is a way without contour integration.
$$I = \int_0^1 \dfrac{dx}{(x^2-x^3)^{1/3}} = \int_0^1 \dfrac{dx}{x^{2/3}(1-x)^{1/3}}$$
Let $x=\sin^2(a)$. We then get that
$$I = \int_0^{\pi/2} \dfrac{2 \sin(a) \cos(a) da}{\sin^{4/3}(a) \cos^{2/3}(a)} = 2 \int_0^{\pi/2} \sin^{-1/3}(a) \cos^{1/3}(a) da = \beta(1/3,2/3) = \dfrac{2 \pi}{\sqrt3}$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\large\tt\mbox{Just another complex contour integration}}$:

\begin{align}&\color{#c00000}{\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}}
=\int_{0}^{1}\pars{{1 \over x} - 1}^{2/3}\,{\dd x \over 1 - x}
=\int_{\infty}^{1}{\pars{x - 1}^{2/3} \over 1 - 1/x}\,\pars{-\,{\dd x \over x^{2}}}
\\[3mm]&=\int_{1}^{\infty}{\pars{x - 1}^{2/3} \over \pars{x - 1}x}\,\dd x
=\color{#c00000}{\int_{0}^{\infty}{x^{2/3} \over x\pars{x + 1}}\,\dd x}
=2\pi\ic\,{\expo{2\pi\ic/3} \over -1}
-\int_{\infty}^{0}{x^{2/3}\expo{4\pi\ic/3} \over x\pars{x + 1}}\,\dd x
\\[3mm]&=-2\pi\ic\,\expo{2\pi\ic/3} + \expo{4\pi\ic/3}
\color{#c00000}{\int_{0}^{\infty}{x^{2/3} \over x\pars{x + 1}}\,\dd x}
\end{align}

\begin{align}&\color{#c00000}{%
\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}}
=\color{#c00000}{\int_{0}^{\infty}{x^{2/3} \over x\pars{x + 1}}\,\dd x}
=-2\pi\ic\,{\expo{2\pi\ic/3} \over 1 - \expo{4\pi\ic/3}}
=\pi\,{-2\ic \over \expo{-2\pi\ic/3} - \expo{2\pi\ic/3}}
\\[3mm]&={\pi \over \sin\pars{2\pi/3}}
\end{align}

$$
\color{#66f}{\large%
\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}
={2\pi \over \root{3}}} \approx {\tt 3.6276}
$$

