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)
 A: There is a completely elementary way to solve this. In the end, I do not see how to find the elementary primitive in a simple and intuitive way (if others do, then please edit the answer accordingly), but Rubi helped me. For this reason I post this but make it cw. I also would like to thank @JanG who pointed me to the question and who actually was the one doing the first changes of variables.
Set
$$
I=\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x(1-x)}\,dx.
$$
By doing $x\mapsto 1-x$ and adding, others have found that
$$
I= \frac{1}{2}\int_0^1\frac{x^{-1/3}(1-x)^{-1/3}}{1-x(1-x)}\,dx.
$$
Next, let $x=(1+y)/2$. Then the integral becomes
$$
I=2^{2/3}\int_{-1}^1\frac{1}{(1-y^2)^{1/3}(3+y^2)}\,dy=
2^{5/3}\int_{0}^1\frac{1}{(1-y^2)^{1/3}(3+y^2)}\,dy.
$$
This is very similar to (and just a $y=\cos t$ away from) the integral @Zacky observes in a comment to the question.
This can be put into Rubi, and surprisinlgy the result is elementary,
$$
I=\biggl[\frac{1}{\sqrt{3}}\arctan\Bigl(\frac{\sqrt{3}}{y}\Bigr)
+\frac{1}{\sqrt{3}}\arctan\Bigl(\frac{\sqrt{3}\bigl(1-(2-2y^2)^{1/3}\bigr)}{y}\Bigr)-\frac{1}{3}\text{artanh}\,y+\text{artanh}\,\Bigl(\frac{y}{1+(2-2y^2)^{1/3}}\Bigr)\biggr]_0^1
$$
Inserting the boundarys, the upper one gives (using a limit) $2\pi/(3\sqrt{3})$ and the lower one gives $0$. Hence
$$
I=\frac{2\pi}{3\sqrt{3}}.
$$
A: 
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
$$\int_0^1 \frac{x^{2/3}(1-x)^{-1/3}}{x^2-x+1}\,dx=\frac{2\pi }{3\sqrt 3}$$
A: The solution heavily exploits symmetry of the integrand. 
Let $$I = \int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2} dx $$
Replace $x$ by $1-x$ and sum up gives
$$\tag{1} 2I = \int_0^1 \frac{x^{2/3}(1-x)^{-1/3} + (1-x)^{2/3}x^{-1/3}}{1-x+x^2} dx = \int_0^1 \frac{x^{-1/3}(1-x)^{-1/3}}{1-x+x^2} dx$$

Let $\ln_1$ be complex logarithm with branch cut at positive real axis, while $\ln_2$ be the one whose cut is at negative real axis. Denote 
$$f(z) = \frac{2}{3}\ln_1(x) - \frac{1}{3}\ln_2 (1-x)$$
Then $f(z)$ is discontinuous along the positive axis, but have different jump in $\arg$ across intervals $[0,1]$ and $[1,\infty)$.
Now integrate $g(z) = e^{f(z)}/(1-z+z^2)$ using keyhole contour. Let $\gamma_1$ be path slightly above $[0,1]$, $\gamma_4$ below. $\gamma_2$ be path slightly above $[1,\infty)$, $\gamma_3$ below. It is easily checked that
$$\int_{\gamma 1} g(z) dz = I \qquad \qquad \int_{\gamma 4} g(z) dz = I e^{4\pi i/3}$$
$$\int_{\gamma 2} g(z) dz = e^{\pi i/3}  \underbrace{\int_1^\infty \frac{x^{2/3}(x-1)^{-1/3}}{1-x+x^2} dx}_J\qquad \int_{\gamma 3} g(z) dz = e^{\pi i} J$$
If we perform $x\mapsto 1/x$ on $J$, we get $\int_0^1 x^{-1/3}(1-x)^{-1/3}/(1-x+x^2)dx$, thus $J = 2I$ by $(1)$.
Therefore $$I(1-e^{4\pi i/3}) + 2I(e^{\pi i / 3} - e^{\pi i}) = 2\pi i\times \text{Sum of residues of } g(z) \text{ at } e^{\pm 2\pi i /3}$$
From which I believe you can work out the value of $I$.
