Evaluating $\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}dx$ 
How can we prove $$\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}\mathrm{d} x=\frac{2\pi}{3\sqrt 3}?$$

Thought 1
It cannot be solved by using contour integration directly. If we replace $-1/3$ with $-2/3$ or $1/3$ or something else, we can use contour integration directly to solve it.
Thought 2
I have tried substitution $x=t^3$ and $x=1-t$. None of them worked. But I noticed that the form of $1-x+x^2$ does not change while applying $x=1-t$.
Thought 3
Recall the integral representation of $_2F_1$ function, I was able to convert it into a formula with $_2F_1\left(2/3,1;4/3; e^{\pi i/3}\right)$ involved. But I think it will only make the integral more "complex". Moreover, I prefer a elementary approach. (But I also appreciate hypergeometric approach)


Here we piggy back off the solution posted by @pisco, organize the analysis with detail on the definitions of $\arg(z)$ and $\arg(1-z)$, and finish by evaluating the resiudes enclosed by the closed "keyhole contour."


Let $f(z)$ be the function given by 
$$f(z)=\frac{z^{2/3}(1-z)^{-1/3}}{z^2-z+1}$$
where choose the branch cut from $0$ to $\infty$ along the positive real axis such that 
$$\arg(z)=\begin{cases}
0&, z=x+i0^+\\\\
2\pi&,z=x+i0^-
\end{cases}$$ 
and we choose the branch cut from $1$ to $\infty$ along the positive real axis with $\arg(1-z)=-\pi+\arg(z-1)$ such that 
$$\arg(1-z)=\begin{cases}
0&, 0<x<1\\\\
-\pi&,z=x+i0^+, 1<x\\\\
\pi&, z=x+i0^-, 1<x
\end{cases}$$ 

Then, the integral around the classical "key hole" contour $C$ is 
$$\begin{align}
\oint_C f(z)\,dz &=(e^{i2(0)/3}e^{-i(0)/3}-e^{i2(2\pi)/3}e^{-i(0)/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\\\\
&+(e^{i2(0)/3}e^{-i(-\pi)/3}-e^{i2(2\pi)/3}e^{-i(\pi)/3})\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{x^2-x+1}\,dx\\\\
&=(1+e^{i\pi/3})\left(\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx+\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{x^2-x+1}\,dx\right)\tag1
\end{align}$$ 

Enforcing the substitution $x\mapsto 1/x$ in the second integral on the right-hand side of $(1)$ reveals
$$\begin{align}
\oint_C f(z)\,dz &=(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}+x^{-1/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\tag2
\end{align}$$ 

Using the identity $x^{2/3}(1-x)^{-1/3}+x^{-1/3}(1-x)^{2/3}=x^{-1/3}(1-x)^{-1/3}$ and observing that $x^2-x+1=(1-x)^2-(1-x)+1$ we find from $(2)$ that 
$$\begin{align}
\oint_C f(z)\,dz &=3(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\\\\
&=3(1+e^{i\pi/3})\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx\tag3
\end{align}$$ 

From the residue theorem we have 
$$\begin{align}
\oint_C f(z)\,dz&=2\pi i \left(\text{Res}\left(f(z), z=\frac12+i\frac{\sqrt3}2\right)+\text{Res}\left(f(z), z=\frac12-i\frac{\sqrt3}2\right)\right)\\\\
&=2\pi i \left(\frac{e^{i2\pi/9}e^{i\pi/9}}{i2\sqrt 3}+\frac{e^{i10\pi/9}e^{-i\pi/9}}{-i2\sqrt 3}\right)\\\\
&=\frac{2\pi}{\sqrt3} (1+e^{i\pi/3})\tag4
\end{align}$$

Finally, setting $(3)$ and $(4)$ equal yields the coveted result
$$\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx=\frac{2\pi }{3\sqrt 3}$$
