<s>[INST] Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ in order to write
$$
I\propto \int_0^1 \sum_{n=0}^\infty y^n 
$$
however I do not know how to manipulate the $\log^2 x$ term when doing this procedure when doing this substitution.  If we can do this the integral would be trivial from here.
Complex methods are okay also, if you want to use this method we have complex roots at $x=(-1)^{1/3}$.  But what contour can we use suitable for the $\log^2x $ term?
Thanks
 [/INST] Here is another way to do it.  Let $$I=\int_{0}^{1} \frac{(\ln(x))^2}{x^2-x+1} \ dx =\int_{0}^{1} \frac{(\ln(x))^2 (1+x)}{1+x^3} \ dx.$$ By a change of variables $x=\frac{1}{u}, \ dx =-\frac{1}{u^2} \ du,$ we have 
$$I=\int_{1}^{\infty} \frac{(\ln(u))^2 (1+u)}{1+u^3} \ du,$$ which implies
$$I=\frac{1}{2} \int_{0}^{\infty} \frac{(\ln(x))^2 (1+x)}{1+x^3} \ dx.$$ Now split the integrand into two terms 
$$I=\frac{1}{2} \int_{0}^{\infty} \frac{(\ln(x))^2}{x^3+1} \ dx + \frac{1}{2} \int_{0}^{\infty} \frac{x(\ln(x))^2}{x^3+1} \ dx.$$ The two terms are equal to one another, which can be shown by the same change of variables $x= \frac{1}{u}$ on the first term.Thus, 
$$I= \int_{0}^{\infty} \frac{(\ln(x))^2}{x^3+1} \ dx.$$ 
Now consider the triple integral 
$$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2)(1+y^2)(1+x^2y^2z^3)} \ dz \ dy \ dx.$$
Make the change of variables $z=\frac{u}{(xy)^{\frac{2}{3}}}, \ dz = \frac{1}{xy^{\frac{2}{3}}} \ du,$ to see 
$$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{(xy)^{\frac{1}{3}}}{(1+x^2)(1+y^2)(1+u^3)} \ du \ dy \ dx.$$ Apply the well known formula 
$$\int_{0}^{\infty} \frac{t^m}{1+t^n} \ dt= \frac{\pi}{n} \csc \left( \frac{\pi(m+1)}{n} \right)$$ thrice to find $$J=\frac{2 \pi^3}{9\sqrt{3}}.$$ 
Now reverse the order of integration as such. $$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2)(1+y^2)(1+x^2y^2z^3)} \ dx \ dz \ dy.$$ Applying partial fractions or Mathematica gives us that 
$$J=\int_{0}^{\infty}\int_{0}^{\infty}\frac{y\ln(y)}{(1+y^2)(-1+y^2z^3)} \ dz \ dy+ \frac{3}{2} \int_{0}^{\infty}\int_{0}^{\infty}\frac{y\ln(z)}{(1+y^2)(-1+y^2z^3)} \ dz \ dy=J_1+J_2$$ Reverse the order of integration on $J_2$ and use partial fractions to see 
$$J_2=\frac{9}{4}I.$$
Now we focus on the first term $J_1$ Apply partial fractions or use Mathematica to see that 
$$J_1=-\frac{\pi}{3 \sqrt{3}} \int_{0}^{\infty} \frac{y^{\frac{1}{3}} \ln(y)}{y^2+1} \ dy.$$
We expand $$J_1 =\frac{\pi}{3 \sqrt{3}} \int_{0}^{\infty} \int_{0}^{\infty} \frac{ty^{\frac{1}{3}}}{(1-t^2y^2)(1+t^2)} \ dt \ dy.$$ This equality is true by using partial fractions to evaluate the inner integral. Reversing the order of integration, and using partial fractions again, we get that 
$$J_1=\frac{-\pi^2}{18} \int_{0}^{\infty} \frac{t^{\frac{-1}{3}}}{1+t^2} \ dt= \frac{-\pi^3}{18\sqrt{3}}$$ by our aforementioned well-known integral formula. 
Since $$J=J_1+ \frac{9}{4}I,$$ and $J=\frac{2 \pi ^3}{9\sqrt{3}}$ and $J_1= \frac{-\pi^3}{18\sqrt{3}},$ we have that $$I=\frac{4}{9}\left(\frac{2 \pi ^3}{9\sqrt{3}}+\frac{\pi^3}{18\sqrt{3}}\right)=\frac{10 \pi ^3}{81\sqrt{3}}.$$
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