<s>[INST] Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ In order to compute, in an elementary way, 
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$ 
(see  Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$) 
i need to show, in a simple way, that:
$\displaystyle \int_0^1 \dfrac{\arctan x \log x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$
$G$ is the Catalan's constant.
 [/INST] I finally get a solution (i swear i didn't know it when i have posted the question)
Define for $x\in [0,1]$ the function $F$:
$\displaystyle F(x)=\int_0^x \dfrac{\ln t}{1+t}dt$
Notice that $F(1)=-\dfrac{\pi^2}{12}$ 
(use Taylor's development)
and, after performing the change of variable $y=\dfrac{t}{x}$,
$\displaystyle F(x)=\int_0^1 \dfrac{x\ln(xy)}{1+xy}dy$
Since that:
$\Big[F(x)\arctan x\Big]_0^1=-\dfrac{\pi^3}{48}$
then,
$\displaystyle -\dfrac{\pi^3}{48}=\int_0^1 \dfrac{F(x)}{1+x^2}dx+\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(xy)}{(1+xy)(1+x^2)}dxdy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(x)}{(1+xy)(1+x^2)}dxdy+\int_0^1\int_0^1 \dfrac{x\ln(y)}{(1+xy)(1+x^2)}dxdy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\left[\dfrac{\ln x\ln(1+xy)}{1+x^2}\right]_{y=0}^{y=1} dx+
\displaystyle \int_0^1 \left[-\dfrac{\ln y\ln(1+xy)}{1+y^2}+\dfrac{\ln y\ln(1+x^2)}{2(1+y^2)}+\dfrac{y\ln y\arctan x}{1+y^2}\right]_{x=0}^{x=1}dy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx= \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx-\int_0^1\dfrac{\ln y\ln(1+y)}{1+y^2}dy+\dfrac{\ln 2}{2}\int_0^1 \dfrac{\ln y}{1+y^2}dy+
\dfrac{\pi}{4}\times \int_0^1 \dfrac{y\ln y}{1+y^2}dy$
Using Taylor's development,
$\displaystyle \int_0^1 \dfrac{y\ln y}{1+y^2}dy=-\dfrac{\pi^2}{48}$
And it's well known that, $\displaystyle -G=\int_0^1\dfrac{\ln y}{1+y^2}dy$
Therefore,
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=-\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{192}$
And finally,
$\displaystyle \int_0^1 \dfrac{\arctan x \ln x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$
(I hope there is no mistake, this proof is too wonderful to be true )
NB:
Added, July 2, 2019.
The above computation is the result of "reverse engineering".
I was searching for a way to express $\pi^3$ as in integral.
If you introduce the function, for $x\in [0;1]$,
\begin{align}\displaystyle F(x)&=\int_0^x \dfrac{\ln t}{1+t}dt\\
&=\int_0^1 \dfrac{x\ln(tx)}{1+tx}dt
\end{align}
Observe that,
\begin{align}\frac{\partial F(x)}{\partial x}&=\dfrac{\ln x}{1+x}\\
F(1)&=-\frac{\pi^2}{12}
\end{align}
Then,
\begin{align}-\frac{\pi^3}{48}&=\Big[F(x)\arctan x\Big]_0^1\\
\end{align}
And, 
\begin{align}\frac{\partial F(x)}{\partial x}\arctan x=\frac{\arctan x\ln x}{1+x}\end{align}
Thus, one can apply integration by parts,
\begin{align}\int_0^1 \frac{\arctan x\ln x}{1+x}\,dx&=\int_0^1 \frac{\partial F(x)}{\partial x}\arctan x\,dx\end{align}
and so on,
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