How to evaluate $\int_0^1\frac{\ln x\ln(1+x^2)}{1-x^2}dx$ in an elegant way? How to prove that:

$$\int_0^1\frac{\ln x\ln(1+x^2)}{1-x^2}dx=\frac74\zeta(3)-\frac34\ln2 \zeta(2)-\frac{\pi}{2}G$$

where $\zeta$ is the Riemann zeta function and $G$ is Catalan constant.
I came across this integral while working on evaluating some harmonic series. 
I am tagging "harmonic series" as its pretty related to logarithmic integrals. 
 A: We will start by using the following substitution:
$$\frac{1-x}{1+x}=t\Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=\frac{2}{(1+t)^2}dt$$
$$\Rightarrow I=\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x^2}dx=\frac12\int_0^1 \frac{[\ln(1-t)-\ln(1+t)][\ln2+\ln(1+t^2)-2\ln(1+t)]}{t}dt$$
Now we are going to use the following result to evaluate a part from above:
$$\small  \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$
$$\Rightarrow I=\frac{7}{8}\zeta(3)-\frac34\zeta(2)\ln 2+\frac12{\int_0^1 \frac{[\ln(1-t)-\ln(1+t)]\ln(1+t^2)}{t}dt}$$
The last integral is $I-J=\frac74\zeta(3)-\pi G$ which appears in the following post.
$$\Rightarrow I =\frac{7}{8}\zeta(3)-\frac34\zeta(2)\ln 2+\frac78\zeta(3)-\frac{\pi}{2}G=\frac74\zeta(3)-\frac34\zeta(2)\ln 2-\frac{\pi}{2}G$$
A: @LeBlanc proved here
$$I=\Im\int_0^1 \frac{\operatorname{Li}_2(ix)}{1+x^2}dx=\frac{7}{8}\zeta(3)-\frac{\pi}{4}G\tag1$$
on the other hand and by using $\operatorname{Li}_2(y)=-\int_0^1\frac{y\ln u}{1-yu}du$, we can write
\begin{align}
I&=-\Im\int_0^1\frac{1}{1+x^2}\left(\int_0^1\frac{ix\ln u}{1-ixu}\ du\right)\ dx\\
&=-\Im\int_0^1\ln u\left(\int_0^1\frac{ix}{(1+x^2)(1-ixu)}\ dx\right)\ du\\
&=-\Im\int_0^1\ln u\left(\frac{i\ln2}{2}\frac{1}{1-u^2}-\frac{i\ln(1-iu)}{1-u^2}+\frac{\pi}{4}\frac{u}{1-u^2}\right)\ du\\
&=\frac12\int_0^1\frac{\ln u\ln(1+u^2)}{1-u^2}du-\frac{\ln2}{2}\underbrace{\int_0^1\frac{\ln u}{1-u^2}du}_{-\frac34\zeta(2)}\tag2
\end{align} 
From (1) and (2) we get the closed form of our integral.
Note that what I did in the second last line is I ignored the last term $\frac{\pi}{4}\frac{u}{1-u^2}$ as we are interested in only the imaginary parts and I used $\Re \ln(1-iu)=\ln\sqrt{1+u^2}$.
