Show that $\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx = \frac{8 \pi ^3}{81 \sqrt{3}}$ I have found myself faced with evaluating the following integral: $$\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx. $$
Mathematica gives a closed form of $8 \pi ^3/(81 \sqrt{3})$, but I have no idea how to arrive at this closed form. I've tried playing around with some methods from complex analysis, but I haven't had much luck (it has been a while). Does anyone have any ideas? Thanks in advance!

Note
$$ \int_0^1x^m(\ln x)^2dx=\frac{2}{(m+1)^3} $$
So
\begin{eqnarray}
I&=&\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1}dx\\
&=&\int_0^1\frac{(1-x)(\ln x)^2}{1-x^3}dx\\
&=&\int_0^1\sum_{n=0}^\infty(1-x)x^{3n}(\ln x)^2dx\\
&=&2\sum_{n=0}^\infty\left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\right)\\
&=&2\sum_{n=-\infty}^\infty\frac{1}{(3n+1)^3}
\end{eqnarray}
Note
\begin{eqnarray}
\sum_{n=-\infty}^\infty\frac{1}{(3n+1)^3}=-\pi\text{Res}(\frac{1}{(z+1)^3}\cot(\pi z),-\frac{1}{3})=\frac{4\pi^3}{81\sqrt3}
\end{eqnarray}
and hence
$$ I=\frac{8\pi^3}{81\sqrt3}. $$
