\begin{align}I&=\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx\\
&=\Big[\ln x\ln(1-x)\ln(1+x^2)\Big]_0^1+\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx-\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx\\
&=\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx-\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx
\end{align}
Let $R$ the function defined for $[0;1]$ by,
\begin{align}
R(x)&=\int_0^x \frac{2t\ln t}{1+t^2}\,dt\\
&=\int_0^1 \frac{2tx^2\ln(tx)}{1+t^2x^2}\,dt
\end{align}For $0<A<1$,
\begin{align}\int_0^A \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx&=\Big[R(x)\ln(1-x)\Big]_0^A+\int_0^A \frac{R(x)}{1-x}\,dx\\
&=R(A)\ln(1-A)+\int_0^A \left(\int_0^1\frac{2tx^2\ln(tx)}{(1-x)(1+t^2x^2)}\,dt\right)\,dx\\
&=R(A)\ln(1-A)+\int_0^1 \left(\int_0^A\frac{2tx^2\ln t}{(1-x)(1+t^2x^2)}\,dx\right)\,dt+\\
&\int_0^A \left(\int_0^1\frac{2tx^2\ln x}{(1-x)(1+t^2x^2)}\,dt\right)\,dx\\
&=R(A)\ln(1-A)-\int_0^1 \frac{\ln t\ln(1+A^2t^2)}{(1+t^2)t}\,dt-2\int_0^1\frac{\ln t\arctan t }{1+t^2}\,dt-\\
&2\ln(1-A)\int_0^1 \frac{t\ln t}{1+t^2}\,dt+\int_0^A \frac{\ln x\ln(1+x^2)}{1-x)}\,dx \end{align}
Take the limit at $A=1$,
\begin{align}\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx&=-\int_0^1 \frac{\ln t\ln(1+t^2)}{(1+t^2)t}\,dt-2\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt+\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx\end{align}In the first integral perform the change of variable $y=x^2$,
\begin{align}\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx&=-\frac{1}{4}\int_0^1 \frac{\ln t\ln(1+t)}{(1+t)t}\,dt-2\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt+\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx\\
&=\frac{1}{4}\int_0^1\frac{\ln x\ln(1+x)}{1+x}\,dx-\frac{1}{4}\int_0^1\frac{\ln x\ln(1+x)}{x}\,dx-2\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt+\\
&\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx
\end{align}In the second integral perform integration by parts,
\begin{align}\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx&=\frac{1}{4}\int_0^1\frac{\ln x\ln(1+x)}{1+x}\,dx+\frac{1}{8}\int_0^1\frac{\ln^2 x}{1+x}\,dx-2\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt+\\
&\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx+\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx\\
&=\frac{1}{4}\int_0^1\frac{\ln x\ln(1+x)}{1+x}\,dx-\frac{1}{8}\int_0^1\frac{2x\ln^2 x}{1-x^2}\,dx+\frac{1}{8}\int_0^1\frac{\ln^2 x}{1-x}\,dx-\\
&2\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt+\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx
\end{align}
In the second integral perform the change of variable $y=x^2$,
\begin{align}\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx&=\frac{1}{4}\int_0^1\frac{\ln x\ln(1+x)}{1+x}\,dx+\frac{3}{32}\int_0^1\frac{\ln^2 x}{1-x}\,dx-\\
&2\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt+\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx\\
&=\frac{1}{4}\int_0^1\frac{\ln x\ln(1+x)}{1+x}\,dx+\frac{3}{16}\zeta(3)-2\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt+\\
&\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}\,dx
\\J&=\int_0^1 \frac{\ln(1+x)\ln x}{1+x}\\
A&=\int_0^1 \frac{\ln^2 x}{1-x^2}\,dx\\
&=\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1 \frac{x\ln^2 x}{1-x^2}\,dx
\end{align}In the latter integral perform the change of variable $y=x^2$:
\begin{align}A&=\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{7}{8}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{7}{4}\zeta(3)
\end{align}On the other hand, perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{align}A&=\frac{1}{2}\int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right) }{x}\,dx\\
B&=\frac{1}{2}\int_0^1 \frac{\ln^2\left(1-x^2\right) }{x}\,dx
\end{align}In the latter integral perform the change of variable $y=1-x^2$,\begin{align}B&=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
B&=\frac{1}{2}\zeta(3)\\
A+B&=\int_0^1 \frac{\ln^2\left(1-x\right) }{x}\,dx+\int_0^1 \frac{\ln^2\left(1+x\right) }{x}\,dx\\
&=\int_0^1 \frac{\ln^2\left(1-x\right) }{x}\,dx+\Big[\ln x\ln(1+x)^2\Big]_0^1-2\int_0^1 \frac{\ln(1+x)\ln x}{1+x}\,dx
\end{align}In the first integral perform the change of variable $y=1-x$,\begin{align}A+B&=\int_0^1 \frac{\ln^2 x}{1-x}\,dx-2J\end{align}But,\begin{align}A+B&=\frac{9}{4}\zeta(3)\end{align}Therefore,\begin{align}J&=\boxed{-\dfrac{1}{8}\zeta(3)}\\
K&=\int_0^1 \frac{\ln x\arctan x}{1+x^2}\,dx\\
2K&=\int_0^1 \frac{\ln x\arctan x}{1+x^2}\,dx-\int_1^\infty \frac{\ln x\arctan\left(\frac{1}{x}\right)}{1+x^2}\,dx\\
&=\int_0^\infty \frac{\ln x\arctan x}{1+x^2}\,dx+\frac{\pi}{2}\int_0^1 \frac{\ln x}{1+x^2}\,dx\\
&=\int_0^\infty \frac{\ln x\arctan x}{1+x^2}\,dx-\frac{1}{2}\text{G}\pi
\end{align}
Let $S$ the function defined on $[0;\infty]$ by,
\begin{align}
S(x)&=\int_0^x\frac{\ln t}{1+t^2}\,dt\\
&=\int_0^1\frac{x\ln(tx)}{1+t^2x^2}\,dt
\end{align}Observe that,
\begin{align}S(0)&=0,\lim_{x\rightarrow \infty} S(x)=0\\
\int_0^\infty \frac{\ln x\arctan x}{1+x^2}\,dx&=\Big[S(x)\arctan x\Big]_0^\infty-\int_0^\infty \frac{S(x)}{1+x^2}\,dx\\
&=-\int_0^\infty\left(\int_0^1 \frac{x\ln(tx)}{(1+x^2)(1+t^2x^2)}\,dt\right)\,dx\\
&=-\int_0^1\left(\int_0^\infty \frac{x\ln t}{(1+x^2)(1+t^2x^2)}dx\right)dt-\int_0^\infty\left(\int_0^1 \frac{x\ln x}{(1+x^2)(1+t^2x^2)}dt\right)dx\\
&=A-\int_0^\infty \frac{\ln x\arctan x}{1+x^2}\,dx
\end{align}
Therefore,\begin{align}
\int_0^\infty \frac{\ln x\arctan x}{1+x^2}\,dx&=\frac{7}{8}\zeta(3)\\
K&=\boxed{\frac{7}{16}\zeta(3)-\frac{1}{4}\text{G}\pi}\\
\int_0^1 \frac{2x\ln x\ln(1-x)}{1+x^2}\,dx&=\frac{1}{2}\text{G}\pi-\frac{23}{32}\zeta(3)+\int_0^1\frac{\ln x\ln(1+x^2)}{1-x}\,dx\\
I&=\boxed{\frac{23}{32}\zeta(3)-\frac{1}{2}\text{G}\pi}
\end{align}
NB: I assume,
\begin{align}\int_0^1 \frac{\ln^2 x}{1-x}\,dx=2\zeta(3)\end{align}
I have computed $\displaystyle \int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx$ using only univariate changes of variable and performing integration by parts.
PS: $\text{I}$ is linked to Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ see:
https://math.stackexchange.com/a/2716753/186817