<s>[INST] Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$ Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
 [/INST] @RonGordon last sum becomes:
\begin{align}
&\sum_{k = 1}^{\infty}\left\{
1 + \left(k + 1\right)^{2}\log\left(1 - {1 \over \left[k + 1\right]^2}\right) \right\}=
\sum_{k = 1}^{\infty}\left\{
1 - \left(k + 1\right)^{2}
\int_{0}^{1}{{\rm d}x \over \left(k + 1\right)^{2} - x}\right\}
\\[3mm]&=
-\left[\int_{0}^{1}\sum_{k = 1}^{\infty}{1 \over \left(k + 1\right)^{2} - x}\right]
\,x\,{\rm d}x
=
-\left[\int_{0}^{1}\sum_{k = 0}^{\infty}
{1 \over \left(k + 2 + x^{1/2}\right)\left(k + 2 - x^{1/2}\right)}\right]
\,x\,{\rm d}x
\\[3mm]&=
\int_{0}^{1}{
\Psi\left(2 - x^{1/2}\right) - \Psi\left(2 + x^{1/2}\right) \over 2x^{1/2}}\,x
\,{\rm d}x
=
\underbrace{\int_{0}^{1}
\color{#c00000}{\left\lbrack\Psi\left(2 - x\right) - \Psi\left(2 + x\right)\right\rbrack}\,x^{2}\,{\rm d}x}
_{\displaystyle{\log\left(\pi\right) - {3 \over 2}}}
\end{align}

since
  \begin{align}
&\color{#c00000}{\Psi\left(2 - x\right) - \Psi\left(2 + x\right)}
=\left[\Psi\left(1 - x\right) + {1 \over 1 - x}\right]
-
\left[\Psi\left(x\right) + {1 \over 1 + x} + {1 \over x}\right]
\\[3mm]&=\left[\Psi\left(1 - x\right) - \Psi\left(x\right) - {1 \over x}\right]
+{2x \over 1 - x^{2}}
=\color{#c00000}{\left[\pi\cot\left(\pi x\right) - {1 \over x}\right]
+ {2x \over 1 - x^{2}}}
\end{align}

Sorry. It was too long for a comment !!!
</s>