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!

It can be observed that $x^{2} + x + 1 = (x-a)(x-b)$ where $a = e^{2\pi i/3}$ and $b = e^{-2\pi i/3}$. Now
\begin{align}
I &= \int_{1}^{\infty} \frac{ (\ln(x))^{2} }{ (x-a)(x-b) } \, dx 
= \frac{1}{a-b} \, \int_{1}^{\infty} \left( \frac{1}{x-a} - \frac{1}{x-b} \right) \, (\ln(x))^{2} \, dx.
\end{align}
From Wolfram Alpha the integral
\begin{align}
\int \frac{ (\ln(x))^{2} }{ x - a } dx = -2 Li_{3}\left( \frac{x}{a} \right) +2 \log(x) \, Li_{2} \left( \frac{x}{a} \right) + \log^{2}(x) \log\left( 1-\frac{x}{a} \right)
\end{align}
for which the integral in question becomes
\begin{align}
I &= \left[ \frac{-2}{a-b} \left(Li_{3}\left( \frac{x}{a} \right) - Li_{3}\left(\frac{x}{b} \right) \right) + \frac{2}{a-b} \log(x) \, \left(Li_{2} \left( \frac{x}{a} \right) - Li_{2}\left( \frac{x}{b} \right) \right) + \frac{1}{a-b} \, \log^{2}(x) \log\left( \frac{a-x}{b-x} \right) \right]_{1}^{\infty} \\
&= \frac{-2}{a-b} \left[ Li_{3}\left( \frac{1}{a} \right) - Li_{3}\left(\frac{1}{b} \right) \right].
\end{align}
This can then be seen as
\begin{align}
I &= \frac{-2i}{\sqrt{3}} \left[ Li_{3}\left( e^{2\pi i/3} \right) - Li_{3}\left( e^{- 2\pi i/3} \right) \right] \\
&= \frac{-2 i}{\sqrt{3}} \cdot \frac{4 \pi^{3} i}{81} = \frac{8 \pi^{3}}{81 \sqrt{3}}. 
\end{align}
\begin{align}
\int_{1}^{\infty} \frac{ (\ln(x))^{2} }{ x^{2} + x + 1 } \, dx = \frac{8 \pi^{3}}{81 \sqrt{3}}.  
\end{align}
