Empirically, i have obtained the following value: \begin{align}K&=\int_0^1 \frac{\arctan x\ln^2 x}{1+x^2}\,dx\\ &=\frac{151}{11520}\pi^4-\frac{1}{24}\ln^4 2-\text{Li}_4\left(\frac{1}{2}\right)+\frac{1}{24}\pi^2\ln^2 2-\frac{7}{8}\zeta(3)\ln 2\end{align}
How to prove this?
My attempt:
Observe:
\begin{align}K&=\int_0^1 \int_0^1\frac{x\ln^2 x}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\
\end{align}
On the other hand,
\begin{align}K&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^2 t}{1+t^2}\,dt\right)\arctan x\right]_0^1-\int_0^1 \int_0^1\frac{x\ln(tx)^2}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ &=\frac{\pi^4}{64}-K-\int_0^1\int_0^1 \frac{x\ln^2 t}{(1+x^2)(1+t^2x^2)}\,dt\,dx-2\int_0^1\int_0^1 \frac{x\ln t\ln x}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ \end{align} Moreover, one can prove: \begin{align}\int_0^1 \int_0^1\frac{x\ln^2 t}{(1+x^2)(1+t^2x^2)}\,dt\,dx&=\frac{1}{64}\pi^4-\text{G}^2\end{align}
Unfortunately, $\displaystyle U= \int_0^1\int_0^1 \frac{x\ln t\ln x}{(1+x^2)(1+t^2x^2)}\,dt\,dx$ seems not easier to compute than $K$
Edit: \begin{align}U&=\int_0^1\int_0^1 \frac{x\ln t\ln x}{(1-t^2)(1+x^2)}\,dt\,dx -\int_0^1\int_0^1 \frac{xt^2\ln t\ln x}{(1-t^2)(1+t^2x^2)}\,dt\,dx\\ &=\frac{1}{384}\pi^4-\int_0^1\int_0^1 \frac{xt^2\ln t\ln(tx)}{(1-t^2)(1+t^2x^2)}\,dt\,dx+\int_0^1\int_0^1 \frac{xt^2\ln^2 t}{(1-t^2)(1+t^2x^2)}\,dt\,dx\\ \end{align} The last one is doable and, \begin{align}V&=\int_0^1\int_0^1 \frac{xt^2\ln t\ln(tx)}{(1-t^2)(1+t^2x^2)}\,dt\,dx\\ &=\int_0^1 \frac{\ln t}{1-t^2}\left(\int_0^t \frac{u\ln u}{1+u^2}\,du\right)\,dt\\ &=\frac{1}{4}\int_0^1 \frac{\ln t}{1-t^2}\left(\int_0^{t^2} \frac{\ln u}{1+u}\,du\right)\,dt\\ \end{align}
Edit2:
Since for $t\neq 1$, $\displaystyle \frac{1}{1-t^2}=\frac{1}{2}\times \frac{2t}{1-t^2}+\frac{1}{1+t}$ then,
\begin{align}V&=\frac{1}{4}\int_0^1 \left(\frac{1}{2}\times \frac{2t}{1-t^2}+\frac{1}{1+t}\right)\ln t\left(\int_0^{t^2} \frac{\ln u}{1+u} \,du\right)\,dt\\ &=\frac{1}{4}\int_0^1 \frac{\ln t}{1+t}\left(\int_0^{t^2} \frac{\ln u}{1+u}\,du\right)\,dt+\frac{1}{16}\int_0^1 \frac{\ln t}{1-t}\left(\int_0^t \frac{\ln u}{1+u}\,du\right)\,dt \end{align}
Different approach
From here we have
$$\frac{\arctan x}{1+x^2}=\frac12\sum_{n=1}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$$
multiply both sides by $\ln^2x$ then integrate from $x=0$ to $x=1$ to get
$$\int_0^1\frac{\arctan x\ln^2x}{1+x^2}dx=\frac12\sum_{n=1}^\infty(-1)^n(H_n-2H_{2n})\int_0^1x^{2n-1}\ln^2x\ dx$$ $$=\sum_{n=1}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n)^3}=\frac18\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}-2\sum_{n=1}^\infty(-1)^n\frac{H_{2n}}{(2n)^3}$$
$$=\frac18\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}-2\Re\sum_{n=1}^\infty(i)^n\frac{H_n}{n^3}$$
where $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}$$=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$
and $\sum_{n=1}^\infty(i)^n\frac{H_n}{n^3}$ can be evaluated using the generating function
\begin{align} \sum_{n=1}^\infty\frac{H_n}{n^3}y^n&=\operatorname{Li}_4\left(\frac{y}{y-1}\right)-\frac12\operatorname{Li}_2^2\left(\frac{y}{y-1}\right)+2\operatorname{Li}_4(y)-\operatorname{Li}_4(1-y)-\ln(1-y)\operatorname{Li}_3(y)\\ &\quad +\frac12\ln^2(1-y)\operatorname{Li}_2(y)+\frac12\operatorname{Li}_2^2(y)+\frac16\ln^4(1-y)-\frac16\ln y\ln^3(1-y)\\ &\quad+\frac12\zeta(2)\ln^2(1-y)+\zeta(3)\ln(1-y)+\zeta(4) \end{align}
now set $y=i$ and consider the real part.