How to find $I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx$ $$I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx=-\frac{\pi^4}{512}+\frac{3\pi^2}{128}\ln^22+\frac{\pi}{8}G\ln2-\frac{21}{64}\zeta(3)\ln2$$
 This integral was proposed to me by a friend, but without the solution.
 I tried the integration by parts, but too complex.
 I tried to find a closed form but without result.   $$I=\int_0^1\frac{\arctan^2x}{1±x}\ln{x}dx$$
 A: Continuing on @AliShather's argument, let's focus on the integral 
\begin{align*}
I_1&=\int_0^1\frac{\arctan^2 x\ln x}{1-x^2}dx\\
&\overset{\rm IBP}=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}+2\underbrace{\frac{\arctan x\mathop{\rm artanh} x \ln x}{1+x^2}}_{x\mapsto \frac{1-x}{1+x}}dx\\
&=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}-2{\frac{\left(\arctan x-\frac{\pi}{4}\right)\mathop{\rm artanh} x \ln x}{1+x^2}}dx\\
&=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}dx-\frac{\pi}{4}\int_0^1\frac{\mathop{\rm artanh} x\ln x}{1+x^2}dx\\
&=-A-\frac{\pi}{4}B
\end{align*}
where $\mathop{\rm artanh}x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$. 
By converting into multiple integrals then integrating back, 
\begin{align*}
A&=\int_{[0,1]^4}\frac{x^2}{\left(1-s^2x^2\right)\left(1+t^2x^2\right)\left(1+u^2x^2\right)}dV\\[3px]
&=\mbox{*Partial fractions*}\\[3px]
&=-\int_0^1\frac{\mathop{\rm artanh}x}{x}\left(2\arctan x-\arctan\frac{1}{x}\right)\arctan\frac{1}{x}\,dx\\
&=-\frac{\pi^4}{64}+\pi\underbrace{\int_0^1\frac{\arctan x \mathop{\rm artanh}x}{x}dx}_{J}
\end{align*}
where we used the identity $\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$. 
By $x\mapsto\frac{1-x}{1+x}$, 
$$J=\int_0^1\frac{\left(\arctan x-\frac\pi 4\right)\ln x}{1-x^2}dx=\frac{\pi^3}{32}+\int_0^1\frac{\arctan x\ln x}{1-x^2}dx$$
Then by using IBP directly to J,
\begin{align*}
J&=-\int_0^1\frac{\arctan x\ln x}{1-x^2}+\frac{\mathop{\rm artanh}x \ln x}{1+x^2}dx\\&=\frac{\pi^3}{64}-\frac{1}{2}\underbrace{\int_0^1\frac{\mathop{\rm artanh}x \ln x}{1+x^2}dx}_B 
\end{align*}
By here, 
$$I_1=\frac\pi 4 B=\frac{3\pi^4}{256}-\frac{G\pi\ln 2}{4}-\frac \pi 2\mathfrak{I}\mathop{\rm Li_3}\frac{1+i}{2}+\frac{\pi^2\ln^2 2}{64}$$
Finally using the closed form of $I_2$ by @AliShather and $I=I_1+I_2$ we should get the desired identity, where of course the closed form of B was not necessary and can be replaced by the partial steps used in the evaluation of B. 
