Prove that $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}$ We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$

Put
\begin{equation*}
f(s) = \int_{0}^{1}\dfrac{x\ln(s+x)}{1+x^2}\, dx.
\end{equation*}
We want to determine $f(1)$. After differentiation we have
\begin{gather*}
f'(s) = \int_{0}^{1}\dfrac{x}{(s+x)(1+x^2)}\, dx = -\int_{0}^{1}\dfrac{s}{(s^2+1)(s+x)}\, dx +\int_{0}^{1}\dfrac{sx+1}{(s^2+1)(x^2+1)}\, dx  =
\\[2ex]
-\dfrac{s\ln(1+s)}{1+s^2}+\dfrac{\ln s}{1+s^2}+\dfrac{s\ln 2 }{2(1+s^2)}+\dfrac{{\pi}}{4(1+s^2)}.
\end{gather*}
Now we integrate wrt $s$ between $0$ and $1$. That yields
\begin{equation*}
f(1)-f(0) = -f(1)+f(0) +\dfrac{\ln^2(2)}{4}+\dfrac{{\pi}^2}{16}.
\end{equation*}
Consequently
\begin{equation*}
f(1)= f(0) +\dfrac{\ln^2(2)}{8} + \dfrac{{\pi}^2}{32}. \tag{1}
\end{equation*}
But
\begin{equation*}
f(0) = \int_{0}^{1}\dfrac{x\ln(x)}{1+x^2}\, dx = \sum_{k=0}^{\infty}\int_{0}^{1}(-1)^kx^{2k+1}\ln x\, dx =\sum_{k=0}^{\infty}\dfrac{(-1)^{k+1}}{4(k+1)^{2}} = -\dfrac{\pi^2}{48}.
\end{equation*}
Finally we substitute that into (1).
\begin{equation*}
f(1) = \dfrac{\ln^2(2)}{8} + \dfrac{{\pi}^2}{96}.
\end{equation*}
