Evaluate by contour integration $\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}$ Evaluate by contour integration [i am learning complex analysis - calculus of residues]
$$\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}$$
I tried by taking $x^3$ out from the denominator but that didnt work. 
 A: Use a dogbone contour with a single branch around [0,1].
Consider $$f(z)=\frac{1}{z^{2/3}(1-z)^{1/3}}$$ and use the branches 
$\displaystyle -\pi\leq arg(z)<\pi, \;\  0\leq arg(1-z)< 2\pi$
Arg above the cut:  $arg(z)=0, \;\ arg(1-z)=2\pi$
Arg below the cut:  $arg(z)=0, \;\ arg(1-z)=0$
$$f(z)=\frac{1}{|z|^{2/3}e^{2i\cdot arg(z)/3}|1-z|^{1/3}e^{i\cdot arg(1-z)/3}}$$
$$\int_{0}^{1}\frac{1}{x^{2/3}e^{2i (0)/3}(1-x)^{1/3}e^{i(2\pi )/3}}dx-\int_{0}^{1}\frac{1}{x^{2/3}e^{2i(0)/3}(1-x)^{1/2}e^{i(0)/3}}dx=2\pi i Res(f,\infty)$$
the residue at infinity is $e^{2\pi i /3}$,  (I have to admit I just done this with Mathematica to save some time, or one may use the Laurent expansion) so:
$$(e^{-2\pi i/3}-1)\int_{0}^{1}\frac{1}{x^{2/3}(1-x)^{1/3}}dx=2\pi i e^{2\pi i/3}$$
$$\int_{0}^{1}\frac{1}{x^{2/3}(1-x)^{1/3}}dx=\frac{2\pi i e^{2\pi i/3}}{e^{-2\pi i/3}-1}=\frac{2\pi}{\sqrt{3}}$$
EDIT:
To find the residue at infinity, one can multiply by $-1/x^{2}$ and let $x\to 1/x$ to obtain the series expansion about 0:
$\displaystyle \frac{-1}{x^{2}}\cdot \frac{1}{\frac{1}{x^{2/3}}(1-\frac{1}{x})^{1/3}}=\frac{e^{2\pi i/3}}{x}+\frac{e^{2\pi i/3}}{3}+\frac{2e^{2\pi i/3}}{9}x+\cdot\cdot\cdot $
The coefficient of the 1/x term is the residue. 
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#c00000}{\int_{0}^{1}{\dd x \over \pars{x^{2} - x^{3}}^{1/3}}}&
=\int_{\infty}^{1}{-\dd x/x^{2} \over \pars{1/x^{2} - 1/x^{3}}^{1/3}}
=\int_{1}^{\infty}{\dd x \over x\pars{x - 1}^{1/3}}
=\color{#c00000}{\int_{0}^{\infty}{x^{-1/3}\,\dd x \over x + 1}}
\end{align}
Use the following contour with
$\ds{z^{-1/3} = \verts{z}^{-1/3}\expo{-\ic\phi\pars{z}/3}\,,\qquad
0 < \phi\pars{z} < 2\pi}$:


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

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

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

