Using a keyhole contour for this specific problem 
I want to compute 
  $$\int\limits_0^1 {1 \over \sqrt[3]{x^2 -x^3}} dx$$  

My book suggests to use a keyhole contour with small circles at branch points $0$ and $1$.  
The answer is ${2 \pi \over \sqrt{3}}$.  
I have tried many things but I do not get an answer, but instead the addition of several integrals, some which go to zero as $\epsilon \to 0$ and $R \to \infty$.
 A: METHODODLGY $1$:  Real Analysis
Note that we can write
$$\begin{align}
\int_0^1 \frac{1}{\sqrt[3]{x^2-x^3}}\,dx&=\int_0^1 x^{-2/3}(1-x)^{-1/3}\,dx\\\\
&=B(1/3,2/3)\tag 1\\\\
&=\frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/3+2/3)}\tag 2\\\\
&=\Gamma(1/3)\Gamma(1-1/3)\tag 3\\\\
&=\frac{\pi}{\sin(\pi/3)}\tag 4\\\\
&=\frac{2\pi}{\sqrt 3}
\end{align}$$
NOTES:
In arriving at $(1)$, we use an integral representation of the Beta function, $B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx$ for $a>0$, $b>0$.
In going from $(1)$ to $(2)$, we use the relationship between the Beta and Gamma functions, $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$
In arriving at $(3)$, we used the fact that $\Gamma(1)=1$.
And in going from $(3)$ to $(4)$, we use the Euler's reflection formula, $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ for non-integer values of $z$.

METHODODLGY $2$:  Complex Analysis
First, we choose to cut the plane with branch cuts from $0$ to $-\infty$ and from $1$ to $-\infty$.  With the plane cut accordingly, the arguments of $z$ and $1-z$ are given by 
$$-\pi<\arg(z)\le \pi$$
and 
$$-2\pi <\arg(1-z)\le 0$$
Moreover, the function $\frac{1}{\sqrt[3]{z^2-z^3}}$ is analytic on $\mathbb{C}\setminus [0,1]$.
Integration over the "dog bone contour" $C$, not the "keyhole" contour, can be written
$$\begin{align}
\oint_{C_\epsilon} \frac{1}{\sqrt[3]{z^2-z^3}}\,dz&=\int_{\epsilon}^{1-\epsilon} \frac{1}{\sqrt[3]{x^2-x^3}}\,dx+\int_{1-\epsilon}^\epsilon \frac{1}{e^{-i2\pi/3}\sqrt[3]{x^2-x^3}}\,dx\\\\
&+\int_0^{2\pi}\frac{i\epsilon e^{i\phi}}{\sqrt[3]{(1+\epsilon e^{i\phi})^2\,(-\epsilon e^{i\phi})}}\,d\phi\\\\
&+\int_{2\pi}^0 \frac{i\epsilon e^{i\phi}}{\sqrt[3]{(\epsilon e^{i\phi})^2}(1-\epsilon e^{i\phi})}\,d\phi\tag 5
\end{align}$$
As $\epsilon \to 0^+$, the third and fourth integrals on the right-hand side of $(5)$ approach $0$, which reveals that
$$\lim_{\epsilon\to 0^+}\oint_{C_\epsilon} \frac{1}{\sqrt[3]{z^2-z^3}}\,dz=(1-e^{i2\pi/3})\int_0^1 \frac{1}{\sqrt[3]{x^2-x^3}}\,dx$$
We can evaluate the contour integral by using Cauchy's Integral Theorem to deform the contour to a circle, centered at $0$, with radius $R$, traversed clockwise.  This yields
$$\begin{align}
\lim_{\epsilon\to 0^+}\oint_{C_\epsilon}\frac{1}{\sqrt[3]{z^2-z^3}}\,dz&=-\lim_{R\to \infty}\int_0^{2\pi}\frac{iRe^{i\phi}}{\sqrt[3]{(Re^{i\phi})^2-(Re^{i\phi})^3}}\,d\phi\\\\
&=-i2\pi \frac{1}{\sqrt[3]{e^{-i\pi}}}\\\\
&=-i2\pi e^{i\pi/3}\tag 6
\end{align}$$
Setting $(5)$ and $(6)$ equal,solving for the integral of interest, and simplifying, we obtain
$$\int_0^1 \frac{1}{\sqrt[3]{x^3-x^3}}\,dx=\frac{2\pi}{\sqrt 3}$$
as expected!
A: Hint:
Use $x=\sin^2t$ and $u=\sqrt[3]{\cot{t}}$.
Finally calculate $6\int\limits_0^{+\infty}\frac{u^3}{u^6+1}du$
A: $$
\begin{align}
\int_0^1\frac1{\sqrt[\large3]{x^2 -x^3}}\,\mathrm{d}x
&=\int_0^1x^{-2/3}(1-x)^{-1/3}\,\mathrm{d}x\tag{1}\\
&=\int_0^\infty\left(\frac{x}{x+1}\right)^{-2/3}\left(\frac1{x+1}\right)^{-1/3}\frac{\mathrm{d}x}{(x+1)^2}\tag{2}\\
&=\int_0^\infty\frac{x^{-2/3}}{1+x}\,\mathrm{d}x\tag{3}\\
&=\frac1{1-e^{-4\pi i/3}}\int_\gamma\frac{z^{-2/3}}{1+z}\,\mathrm{d}z\tag{4}\\
&=\frac{2\pi ie^{-2\pi i/3}}{1-e^{-4\pi i/3}}\tag{5}\\[4pt]
&=\frac\pi{\sin\left(\frac{2\pi}3\right)}\tag{6}\\
&=\frac{2\pi}{\sqrt3}\tag{7}
\end{align}
$$
Explanation:
$(1)$: rewrite the integrand
$(2)$: substitute $x\mapsto\frac{x}{x+1}$
$(3)$: algebra
$(4)$: $\gamma$ is the keyhole contour $$\left[i\epsilon,Re^{i\arcsin\left(\frac\epsilon R\right)}\right]\cup Re^{i\left[\arcsin\left(\frac\epsilon R\right),2\pi-\arcsin\left(\frac\epsilon R\right)\right]}\cup\left[Re^{-i\arcsin\left(\frac\epsilon R\right)},-i\epsilon\right]\cup\epsilon e^{i\left[\frac{3\pi}2,\frac\pi2\right]}$$
$(5)$: the residue of $\frac{z^{-2/3}}{1+z}$ at $z=-1$ is $e^{-2\pi i/3}$
$(6)$: Use Euler's Formula
$(7)$: evaluate
Diagram of $\gamma$ ($R\to\infty$ and $\epsilon\to0$):


Addendum to Mark Viola's Answer
In Mark Viola's answer, it is stated that

Moreover, the function $\frac{1}{\sqrt[3]{z^2-z^3}}$ is analytic on $\mathbb{C}\setminus [0,1]$.

We can write
$$
\begin{align}
\log\left(\sqrt[\large3]{z^2-z^3}\right)
&=\frac13\log\left(z^2-z^3\right)\\
&=\frac{\log(2)}3+\frac13\int_{-1}^z\frac{2w-3w^2}{w^2-w^3}\mathrm{d}w\\
&=\frac{\log(2)}3+\int_{-1}^z\frac13\left(\frac2w+\frac1{w-1}\right)\mathrm{d}w\tag{8}
\end{align}
$$
The sum of the residues at $0$ and $1$ of the integrand in $(8)$ is $1$. This means that if we circle both singularities, we get a total integral of $2\pi i$. Therefore, via exponentiation, $\sqrt[\large3]{z^2-z^3}$ is defined unambiguously by the integral in $(8)$ for all $z\in\mathbb{C}\setminus[0,1]$.
