calculus - Evaluating $int_0^1frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}dx$

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)

Answer

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
$$\tag{1} 2I = \int_0^1 \frac{x^{2/3}(1-x)^{-1/3} + (1-x)^{2/3}x^{-1/3}}{1-x+x^2} dx = \int_0^1 \frac{x^{-1/3}(1-x)^{-1/3}}{1-x+x^2} dx$$

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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 $I$.