Calculate $\int_0^1\frac{\log^2x\log(1+x^2)}{1-x^2}dx$ Evaluating $$\int_0^1\frac{\log^2(x)\log(1+x^2)}{1-x^2}dx$$
I found      $- \dfrac{\pi^4}{32}+2G^2+\dfrac74 ζ(3)\log2 $  where $G$ is the Catalan's constant.
 A: \begin{align}J&=\int_0^1\frac{\log^2(x)\log(1+x^2)}{1-x^2}dx\\
&\overset{\text{IBP}}=\left[\left(\int_0^x\frac{\ln^2 t}{1-t^2}dt\right)\ln(1+x^2)\right]_0^1 -\int_0^1 \frac{\color{red}{2x}}{1+x^2}\left(\int_0^x\frac{\ln^2 t}{1-t^2}dt\right)dx\\
&=\frac{7}{4}\zeta(3)\ln 2-\int_0^1\int_0^1 \frac{2x^2\ln^2(tx)}{(1+x^2)(1-t^2x^2)}dtdx\\
&=\frac{7}{4}\zeta(3)\ln 2+2\int_0^1\int_0^1 \left(\frac{\ln^2(tx)}{(1+t^2)(1+x^2)}-\frac{\ln^2(tx)}{(1+t^2)(1-t^2x^2)}\right)dtdx\\
&=\frac{7}{4}\zeta(3)\ln 2+4\int_0^1\int_0^1 \int_0^1\frac{\ln^2 x}{(1+t^2)(1+x^2)}dtdx+4\left(\int_0^1\frac{\ln x}{1+x^2}dx\right)^2-\\&2\int_0^1\int_0^1\frac{\ln^2(tx)}{(1+t^2)(1-t^2x^2)}dtdx\\
&=\frac{7}{4}\zeta(3)\ln 2+4\times\frac{1}{16}\pi^3\times \frac{1}{4}\pi+4\text{G}^2-\int_0^1 \frac{2}{t(1+t^2)}\left(\int_0^t\frac{\ln^2 u}{1-u^2}du\right)dt\\
&\overset{\text{IBP}}=\frac{7}{4}\zeta(3)\ln 2+\frac{1}{16}\pi^4+4\text{G}^2-\left[\left(2\ln t-\ln(1+t^2)\right)\left(\int_0^t\frac{\ln^2 u}{1-u^2}du\right)\right]_0^1+\\&\int_0^1 \frac{\left(2\ln t-\ln(1+t^2)\right)\ln^2 t}{1-t^2}dt\\
&=\frac{7}{4}\zeta(3)\ln 2+\frac{1}{16}\pi^4+4\text{G}^2+\frac{7}{4}\zeta(3)\ln 2+2\int_0^1 \frac{\ln^3 t}{1-t^2}dt-J\\
J&=\boxed{\frac{7}{4}\zeta(3)\ln 2+2\text{G}^2-\frac{1}{32}\pi^4}
\end{align}
NB: i assume that,
\begin{align}\int_0^1 \frac{\ln t}{1+t^2}dt&=-\text{G}\\
\int_0^1 \frac{\ln^2 t}{1+t^2}dt&=\frac{\pi^3}{16}\\
\int_0^1 \frac{\ln^2 t}{1-t^2}dt&=\frac{7}{4}\zeta(3)\\
\int_0^1 \frac{\ln^3 t}{1-t^2}dt&=-\frac{\pi^4}{16}
\end{align}
Addendum: typo fixed. Thanks  Sewer Keeper.
