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

Note
$$ \frac{1}{x^2-x+1}=\frac{1+x}{1+x^3}. $$
So
\begin{eqnarray*}
I&=&\int_0^1 \frac{\ln^2x}{x^2-x+1}dx=\int_0^1 \frac{(1+x)\ln^2x}{1+x^3}dx\\
&=&\sum_{n=0}^{\infty}(-1)^n \int_0^1(1+x)x^{3n}\ln^2xdx\\
&=&\sum_{n=0}^{\infty}(-1)^n \int_0^1(x^{3n}+x^{3n+1})\ln^2xdx\\
&=&2\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{(3n+1)^3}+\frac{1}{(3n+2)^3}\right)\\
&=&2\sum_{n=-\infty}^{\infty}(-1)^n\frac{1}{(3n+1)^3}\\
&=&\frac{10\pi^3}{81\sqrt3}.
\end{eqnarray*}
Here we use the following theorem
$$ \sum_{n=-\infty}^\infty (-1)^nf(n)=-\pi \sum_{k=1}^m\text{Re}(\frac{f(z)}{\sin\pi z},a_k) $$
where $a_1,a_2,\cdots,a_m$ are poles of $f(z)$. For $f(z)=\frac{1}{(3z+1)^3}$, $z=-\frac{1}{3}$ is the only pole and
$$ \text{Re}(\frac{f(z)}{\sin\pi z},-\frac{1}{3})=-\frac{5\pi^2}{81\sqrt3}. $$
Thus we have the result.
