Integrating $\int_{0}^{\infty} \frac{(\log x)^2}{x^2+x+1}$ using residue theorem Just out of curiosity, how does one integrate something like this using residue theory?

$$\int_{0}^{\infty}\frac{(\log x)^2}{x^2+x+1} dx$$

According to Wolfram Alpha, the answer is $\dfrac{16\pi^3}{81\sqrt{3}}$.
I have seen similar integrals before, like here and here, and they all require seem some sort of ingenuity. I am sure local Ramanujans will come to rescue soon. :)

Hint ;
Define the following function $$f(z)=\frac {\log^3 (z)}{z^2+z+1}$$ on a key-hole contour with branch cut on the positive real axis . The function has poles $z=e^{\frac{2\pi}{3}i},e^{\frac{4\pi}{3}i}$ .Since the function is analytic in and on the contour with the poles inside the contour we can use the residue theorem 
$$\int^{\infty}_0 f(x)\, dx -\int^{\infty}_0 \frac{\left( \log(x)+ 2 \pi i \right)^3}{x^2+x+1}\, dx= 2\pi i \text{Res}\,\left( f(z) ; e^{\frac{2\pi}{3}i},e^{\frac{4\pi}{3}i}\right)$$
Note:We can easily prove that the integration around the small and big circles approaches $0$.
Note: the third power will cancel and the first power is equal to $0$.
