\begin{align} \int^\infty_0\frac{x\ln^2{x}}{x^4+x^2+1}dx &=\frac{1}{8}\int^\infty_0\frac{\ln^2{x}}{x^2+x+1}dx\\\ &=\frac{1}{4}\int^1_0\frac{(1-x)\ln^2{x}}{1-x^3}dx\\\ &=\frac{1}{4}\sum^\infty_{n=0}\int^1_0\left(x^{3n}-x^{3n+1}\right)\ln^2{x}dx\\\ &=\frac{1}{2}\sum^\infty_{n=0}\left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\right)\\\ &=-\frac{1}{2}\operatorname*{Res}_{z=-1/3}\frac{\pi\cot(\pi z)}{(3z+1)^3}\\\ &=-\frac{1}{108}\left(2\pi^3\cot(\pi z)\csc^2(\pi z)\right)\Bigg{|}_{z=-1/3}\\\ &=\frac{2\pi^3}{81\sqrt{3}} \end{align}