Evaluate by contour integration
$$\int_{0}^{1}\frac{dx}{(x^2-x^3)^\frac{1}{3}}$$
I am having trouble selecting the right contour for the problem as the usual problems involve selecting real axis from $-R$ to $+R$ and then closing with a semicircle of radius $R$. but that doesn't work here.
$\displaystyle{\large% {\cal I} \equiv \int_{0}^{1}{{\rm d}x \over \left(x^{2} - x^{3}\right)^{1/3}} = ?}$
Let's rewrite the integral ${\cal I}$ as follows: $$ {\cal I} = \int_{\infty}^{1}{-1/x^{2} \over \left(1/x^{2} - 1/x^{3}\right)^{1/3}}\,{\rm d}x = \int^{\infty}_{1}{1/x \over \left(x - 1\right)^{1/3}}\,{\rm d}x = \int^{\infty}_{0}{1 \over 1 + x}\,{{\rm d}x \over x^{1/3}} $$ $$ ----------------------------------- $$ We integrate $\displaystyle{{1 \over \left(1 + z\right)z^{1/3}}\,,\quad z\ \in\ {\mathbb C}}$ ( see the integration path at the end ). $\displaystyle{z^{1/3}}$ is defined as $\displaystyle{% \left\vert z\right\vert^{1/3}{\rm e}^{{\rm i}\phi\left(z\right)/3}}$ where $\displaystyle{z \not= 0}$ and $\displaystyle{0 < \phi\left(z\right) < 2\pi}$. This is the $z^{1/3}$ branch cut. With this definition, the pole $z = -1$ becomes ${\rm e}^{{\rm i}\pi}$ such that $z^{1/3} = {\rm e}^{{\rm i}\pi/3}$. Similarly, on the path above the $x$ axis it's $x^{1/3}$ and below it's $\left(x{\rm e}^{2\pi{\rm i}}\right)^{1/3}$. The residue at $x = -1$ is given by $\lim_{x \to {\rm e}^{{\rm i}\pi}}{1 + x \over \left(1 + x\right)x^{1/3}} = {1 \over \vphantom{\Large A}{\rm e}^{{\rm i}\pi/3}}$.
$$ 2\pi{\rm i}\,{1 \over {\rm e}^{{\rm i}\pi/3}} = \int^{\infty}_{0}{1 \over 1 + x}\,{{\rm d}x \over x^{1/3}} + \int_{\infty}^{0}{1 \over 1 + x}\, {{\rm d}x \over x^{1/3}\,{\rm e}^{2\pi{\rm i}/3}} = \left(1 - {\rm e}^{-2\pi{\rm i}/3}\right)\int^{\infty}_{0}{1 \over 1 + x}\,{{\rm d}x \over x^{1/3}} $$
$$ {\cal I} = {2\pi{\rm i} \over {\rm e}^{{\rm i}\pi/3}\left(1 - {\rm e}^{-2\pi{\rm i}/3}\right)} = {2\pi{\rm i} \over {\rm e}^{{\rm i}\pi/3} - {\rm e}^{-{\rm i}\pi/3}} = {2\pi{\rm i} \over 2{\rm i}\sin\left(\pi/3\right)} = {\pi \over \sqrt{\vphantom{\large A}3\,}\,/2} $$
The remaining integrals on the semicircles vanish out. The "big one" $\sim R^{-1/3}$ when $R \to \infty$. The "small one" $\sim \epsilon^{2/3}$ when $\epsilon \to 0$. $$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \int_{0}^{1}{{\rm d}x \over \left(x^{2} - x^{3}\right)^{1/3}} \color{#000000}{=} {2 \over 3}\,\sqrt{\vphantom{\large A}3\,}\,\pi\quad} \\ \\ \hline \end{array} $$