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!

Shocked, shocked! that there is no contour integration yet.  So, without further ado...
Note that
$$f(x) = \frac{\log^2{x}}{x^2+x+1} \implies f \left ( \frac1{x} \right ) = x^2 f(x) $$
Thus,
$$\int_1^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \int_0^{1} \frac{\log^2{x}}{x^2+x+1} = \frac12 \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} $$
Now consider 
$$\oint_C dz \frac{\log^3{z}}{z^2+z+1} $$
where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$.  Taking the limit as $R \to \infty$ and $\epsilon \to 0$, we get that the contour integral is equal to
$$\int_0^{\infty} dx \frac{\log^3{x} - (\log{x}+i 2 \pi)^3}{x^2+x+1} $$
or
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{x^2+x+1} +i 8 \pi^3 \int_0^{\infty} dx \frac{1}{x^2+x+1} $$  
Note that the first integral is what we seek, the second integral is zero (by the same trick we applied above), and the third integral is relatively easy to find:
$$\int_0^{\infty} \frac{dx}{x^2+x+1} = \int_0^{\infty} \frac{dx}{(x+1/2)^2+3/4} = \frac{2}{\sqrt{3}} \left [\arctan{\frac{2}{\sqrt{3}} \left ( x+\frac12 \right )} \right ]_0^{\infty} = \frac{2 \pi}{3 \sqrt{3}}$$
The contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand, which are at $z_+ = e^{i 2 \pi/3}$ and $z_- = e^{i 4 \pi/3}$.  The sum of the residues is
$$\frac{-i 8 \pi^3/27}{i \sqrt{3}} + \frac{-i 64 \pi^3/27}{-i \sqrt{3}}  = \frac{56 \pi^3}{27 \sqrt{3}}$$
Then
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = i 2 \pi \frac{56 \pi^3}{27 \sqrt{3}} - i 8 \pi^3 \frac{2 \pi}{3 \sqrt{3}} = -i \frac{32 \pi^4}{27 \sqrt{3}}$$
Thus,
$$\int_1^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \frac12 \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \frac{8 \pi^3}{81 \sqrt{3}} $$
