Prove that $$I(t) = \int_0^1 \frac{x\ln (1+x)}{1+x^2} \, dx= \frac{\pi^2}{96}+\frac{\ln(2)^2}{8}.$$
I tried this and broke it in a form where I can use 2nd order polylogarithm but could not proceed. This answer is confirmed by Wolfram Alpha.
Please help.
For $t>0$, let $$I(t) = \int_0^1 \frac{x\ln (1+tx)}{1+x^2} \, dx$$ then, we have that $$I'(t)=\int_0^1 \frac{x^2}{(1+tx)(1+x^2)} \, dx= \frac{4\ln(1+t)+2\ln(2)t^2-\pi t}{4t(1+t^2)} $$ By noting that $I(0^+)=0$, integrate with respect $t$ in order to compute $I(1)$ $$I(1)=\int_0^1\frac{\ln(1+t)}{t}dt-I(1)+\frac{\ln(2)}{2}\int_0^1\frac{t}{1+t^2}dt-\frac{\pi}{4}\int_0^1\frac{1}{1+t^2}dt$$ and we obtain $$2I(1)=\frac{\pi^2}{12}+\frac{\ln(2)^2}{4}-\frac{\pi^2}{16}\implies I(1)=\frac{\pi^2}{96}+\frac{\ln(2)^2}{8}.$$
P.S. Note that $$\int_0^1\frac{\ln(1+t)}{t}dt=\int_0^1\sum_{n\geq 1}\frac{(-t)^{n-1}}{n}dt=-\sum_{n\geq 1}\frac{(-1)^{n}}{n^2}=-\mbox{Li}_2(-1)=\frac{\pi^2}{12}.$$
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$\ds{I \equiv \int_{0}^{1}{x\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x = {\phantom{^{2}}\pi^{2} \over 96} + {\ln^{2}\pars{2} \over 8}:\ {\large ?}}$.
\begin{align} I & \equiv \int_{0}^{1}{x\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x = \Re\int_{0}^{1}{\ln\pars{1 + x} \over x - \ic}\,\dd x = \Re\int_{-\ic}^{1 - \ic}{\ln\pars{1 + \ic + x} \over x}\,\dd x \\[5mm] & = \Re\bracks{\ln\pars{1 + \ic}\int_{-\ic}^{1 - \ic}{\dd x \over x} + \int_{-\ic}^{1 - \ic}\ln\pars{1 - {x \over -1 - \ic}}\,{\dd x \over x}} \\[5mm] & = \Re\bracks{\ln\pars{1 + \ic}\ln\pars{1 - \ic} - \ln\pars{1 + \ic}\ln\pars{-\ic} + \int_{-\ic/\pars{-1 - \ic}}^{\pars{1 - \ic}/\pars{-1 - \ic}} {\ln\pars{1 - x} \over x}\,\dd x} \\[1cm] & = \overbrace{\verts{\ln\pars{1 + \ic}}^{2}} ^{\ds{{\phantom{^{2}}\pi^{2} \over 16} + {\ln^{2}\pars{2} \over 4}}}\ -\ {\pi \over 2}\,\ \overbrace{\Im\ln\pars{1 + \ic}}^{\ds{\pi \over 4}}\ -\ \underbrace{\overbrace{\Re\mrm{Li}_{2}\pars{\ic}} ^{\ds{\Re\sum_{n = 1}^{\infty}{\ic^{n} \over n^{2}} = -\,{1 \over 8}\sum_{n = 1}^{\infty}{1 \over n^{2}}}}} _{\ds{-\,{\pi^{2} \over 48}}}\ +\ \Re\mrm{Li}_{2}\pars{{1 \over 2} + {1 \over 2}\,\ic} \\[5mm] & = -\,{\pi^{2} \over 24} + {\ln^{2}\pars{2} \over 4} + {1 \over 2}\braces{\mrm{Li}_{2}\pars{{1 \over 2} + {1 \over 2}\,\ic} + \mrm{Li}_{2}\pars{1 - \bracks{{1 \over 2} + {1 \over 2}\,\ic}}} \label{1}\tag{1} \end{align}
The last polyLogarithm expression can be evaluated with Euler Reflection Formula.
Namely, \begin{align} &{1 \over 2}\braces{\mrm{Li}_{2}\pars{{1 \over 2} + {1 \over 2}\,\ic} + \mrm{Li}_{2}\pars{1 - \bracks{{1 \over 2} + {1 \over 2}\,\ic}}} \\[5mm] = &\ {1 \over 2}\bracks{{\phantom{^{2}}\pi^{2} \over 6} - \ln\pars{{1 \over 2} + {1 \over 2}\,\ic} \ln\pars{{1 \over 2} - {1 \over 2}\,\ic}} = {\pi^{2} \over 12} - {1 \over 2}\verts{\ln\pars{{1 \over 2} + {1 \over 2}\,\ic}}^{2} \\[5mm] = &\ {\pi^{2} \over 12} - {1 \over 2}\verts{-\,{1 \over 2}\,\ln\pars{2} + {\pi \over 4}\,\ic}^{2} = {5\pi^{2} \over 96} - {\ln^{2}\pars{2} \over 8} \end{align}