TITLE: Evaluating $\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}dx$
QUESTION [21 upvotes]: 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)

REPLY [7 votes]: 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
$$
I=\frac{2\pi}{3\sqrt{3}}.
$$