Logarithmic Integral I Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{ \ln^{2}x}{(x^{2} - x + 1)^{2}} \, dx.
\end{align}
It is speculated that the value is
\begin{align}
I = \frac{10 \, \pi^{3}}{3^{5} \, \sqrt{3}} - \zeta(2) + \frac{1}{18} \, \left( \psi^{(1)}\left(\frac{1}{3}\right) + \psi^{(1)}\left(\frac{1}{6}\right) \right).
\end{align}
Is it possible to obtain a demonstration of this result?
 A: Write $\dfrac{1}{(x^2-x+1)^2}$ as a Taylor series $\sum_{j=0}^\infty a_j x^j$.  The coefficients can be written as
$$ a_j = \dfrac{j+1}{3} b_j + \dfrac{2}{3} c_j $$
where $b_j$ repeats $1,2,1,-1,-2,-1,\ldots$ for $j = 0,1,2,\ldots$
while $c_j$ repeats $1,1,0,-1,-1,0,\ldots$, both with period $6$.
Now $$\int_0^1 \log^2(x)\; x^k\; dx = \dfrac{2}{(k+1)^3}$$
so your integral becomes
$$ \dfrac{2}{3} \sum_{j=0}^\infty \dfrac{b_j}{(j+1)^2} + \dfrac{4}{3} \sum_{j=0}^\infty \dfrac{c_j}{(j+1)^3} $$
For $j \equiv 0 \mod 6$ we have $b_j = 1$, $c_j = 1$ and the contribution is $$ \eqalign{\dfrac{2}{3} \sum_{i=0}^\infty \dfrac{1}{(6i+1)^2}  + \dfrac{4}{3} \sum_{i=0}^\infty \dfrac{1}{(6i+1)^3} &= \dfrac{1}{54} \sum_{i=0}^\infty \dfrac{1}{(i+1/6)^2} + \dfrac{1}{162} \sum_{i=0}^\infty \dfrac{1}{(i+1/6)^3}\cr & = \dfrac{\psi^{(1)}(1/6)}{54} + \dfrac{\psi^{(2)}(1/6)}{324}} $$
Similarly we can compute the contributions for $j = 1$ to $5 \mod 6$.
I get (with Maple's help) a total of
$$ \dfrac{\pi^2}{162} + 
{\frac {\psi^{(1)} \left(1/6 \right) }{54}}-{\frac {\psi^{(2)} \left(1/6
 \right) }{324}}+\dfrac{\psi^{(1)} \left( 1/3 \right)}{27} -{\frac {\psi^{(2)} \left( 
1/3 \right) }{324}}-{\frac {\psi^{(1)} \left( 
2/3 \right) }{54}}+{\frac {\psi^{(2)} \left( 2/3 \right) }{324}}-\dfrac{
\psi^{(1)} \left(5/6 \right)}{27} +{\frac {\psi^{(2)} \left( 5/6 \right) }{324}}
$$
Using the reflection identities for $\psi$ this becomes
$$
\dfrac{\psi^{(1)}(1/6)}{18} + \dfrac{\psi^{(1)}(1/3)}{18} 
- \dfrac{\pi^2}{6} +{\frac {20\,{\pi }^{3}\sqrt {3}}{729}}
$$
