I was evaluating the integral $$\mathcal{I}=\int_0^1 \frac{\ln^2 x\ln(1-x)}{x^2+1}\mathrm dx$$
I tried to use the series $$\ln(1-x)=-\sum_{n=1}^\infty \frac{x^n}{n}$$ and interchanged the sum and the integral. $$\mathcal{I}=-\sum_{n=1}^\infty \frac{1}{n}\overbrace{\int_0^1 \frac{x^n\cdot \ln^2 x}{x^2+1}\mathrm dx}^{\mathcal{J}}$$ Now, it's quite obvious that $$\mathcal{J}=\frac{\mathrm d^2}{\mathrm dn^2}\int_0^1 \frac{x^n}{x^2+1}\mathrm dx$$ Now, I found in the answer here by @Olivier Oloa that $$\int_0^1 \frac{x^m}{1+x^n}\mathrm dx=\frac{1}{2n}\psi\bigg(\frac{m+n+1}{2n}\bigg)-\frac{1}{2n}\psi\bigg(\frac{m+1}{2n}\bigg)$$. So, \begin{align}\mathcal{J}&=\frac{\mathrm d^2}{dn^2}\bigg[\frac{1}{4}\psi\bigg(\frac{n+3}{4}\bigg)-\frac{1}{4}\psi\bigg(\frac{n+1}{4}\bigg)\bigg]\\&=\frac{1}{64}\bigg(\psi^{(2)}\bigg(\frac{n+3}{4}\bigg)-\psi^{(2)}\bigg(\frac{n+1}{4}\bigg)\bigg)\end{align} So, the integral converts to the scary-looking series $$\boxed{\mathcal{I}=\frac{-1}{64}\sum_{n=1}^\infty \frac{1}{n}\bigg(\psi^{(2)}\bigg(\frac{n+3}{4}\bigg)-\psi^{(2)}\bigg(\frac{n+1}{4}\bigg)\bigg)}$$ Can someone hint on how to evaluate this infinite series? I want a solution that is based on this approach that I gave.