<s>[INST] How to evaluate the integral $\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$ $$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$$
My attempt:
$$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2-2x(1-x)+(x+1-x)^2}}dx$$
$$=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{(1-x(1-x))^2}}dx=\int_{0}^{1} \frac{(\log {x})^2}{(x^2-x+1)}dx$$
How do I proceed further next step?
 [/INST] Here is another method:
\begin{eqnarray}
I&=&\int_{0}^{1} \frac{(\log {x})^2}{x^2-x+1}dx=\int_{0}^{1} \frac{(1+x)(\log {x})^2}{1+x^3}dx\\
&=&\int_{0}^{1}(1+x)\log^2x\sum_{k=0}^\infty(-1)^kx^{3k}dx\\
&=&\sum_{k=0}^\infty(-1)^k\int_0^1(x^{3k}+x^{3k+1})\log^2xdx\\
&=&2\sum_{k=0}^\infty(-1)^k\left[\frac{1}{(3k+1)^3}+\frac{1}{(3k+2)^3}\right]\\
&=&2\sum_{k=-\infty}^\infty(-1)^k\frac{1}{(3k+1)^3}\\
&=&-2\pi\text{Res}\left(\frac{1}{(3z+1)^3\sin(\pi z)},z=-\frac13\right)\\
&=&\frac{10\pi^3}{81\sqrt3}.
\end{eqnarray}
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