<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] Real Part
Substituting $x\mapsto1/x$ says
$$
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
=\int_1^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{1}
$$
Therefore,
$$
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
=\frac12\int_0^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{2}
$$

Contour Integration Part
Putting the branch cut for $\log(z)$ along the positive real axis, and using the contour
$$
\gamma=[0,R]e^{i\epsilon}\cup Re^{i[\epsilon,2\pi-\epsilon]}\cup[R,0]e^{-i\epsilon}\tag{3}
$$
as $R\to\infty$ and $\epsilon\to0$, $\log(z)=\log(x)$ on the outbound segment and $\log(z)=\log(x)+2\pi i$ on the inbound segment and he integral around huge circular arc vanishes.  Therefore,
$$
\begin{align}
\int_\gamma\frac{\log(z)^3}{z^2-z+1}\mathrm{d}z
&=\int_0^\infty\frac{-6\pi i\log(x)^2+12\pi^2\log(x)+8\pi^3i}{x^2-x+1}\mathrm{d}x\\
&=\frac{124\pi^3}{27\sqrt3}\cdot2\pi i\tag{4}
\end{align}
$$
where $\dfrac{124\pi^3}{27\sqrt3}$ is the sum of the residues of $\dfrac{\log(z)^3}{z^2-z+1}$ at $e^{i\pi/3}$ and $e^{i5\pi/3}$.

Combining Real and Complex Analysis
Therefore, using $(2)$ and the imaginary part of $(4)$, we get
$$
\begin{align}
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
&=\frac12\int_0^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\\
&=\frac12\left[-\frac{124\pi^3}{81\sqrt3}
+\frac{4\pi^2}{3}\int_0^\infty\frac1{x^2-x+1}\mathrm{d}x\right]\\
&=\frac12\left[-\frac{124\pi^3}{81\sqrt3}
+\frac{4\pi^2}{3}\frac{4\pi}{3\sqrt3}\right]\\
&=\frac{10\pi^3}{81\sqrt3}\tag{5}
\end{align}
$$
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