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.
 A: Let we deal with a basic problem first, i.e. the computation of
$$ C_{2n+1} = \int_{0}^{1}\frac{x^{2n+1}\log x}{1+x}\,dx = \int_{0}^{+\infty}\frac{t e^{-(2n+2)t}}{1+e^{-t}}\,dt\tag{1}$$
Since $\int_{0}^{+\infty}t e^{-mt}\,dt = \frac{1}{m^2}$, we have:
$$ -C_{2n+1} = \frac{1}{(2n+2)^2}-\frac{1}{(2n+3)^2}+\frac{1}{(2n+4)^2}-\ldots=\frac{\psi'(n+1)-\psi'\left(n+\frac{3}{2}\right)}{4}\tag{2}$$
and:
$$ I=\int_{0}^{1}\frac{\arctan(x)\log(x)}{1+x}\,dx = -\sum_{n\geq 0}\frac{(-1)^n C_{2n+1}}{2n+1}=-\sum_{m\geq 0}\sum_{n\geq 0}\frac{(-1)^{n+m}}{(2n+1)(2n+m+2)^2}\tag{3}$$
By reindexing the last double series,
$$ I = -\sum_{s=0}^{+\infty}\sum_{p=0}^{s}\frac{(-1)^s}{(2p+1)(p+s+2)^2}=-\sum_{p=0}^{+\infty}\sum_{s\geq p}\frac{(-1)^s}{(p+s+2)^2(2p+1)}\tag{4}$$
hence, in terms of the Hurwitz zeta function:
$$ I = -\sum_{p\geq 0}\frac{(-1)^p}{4(p+1)}\left(\zeta\left(2,p+1\right)-\zeta\left(2,p+\frac{3}{2}\right)\right)\tag{5}$$
or, by using the inverse Laplace transform:
$$ I = -\int_{0}^{+\infty}\frac{s e^{s/2}\log(1+e^{-s})}{4(1+e^{s/2})}\,ds =-\int_{0}^{+\infty}\frac{s e^s \log(1+e^{-2s})}{1+e^s}\,ds\tag{6}$$
where the last integral is a bit more manageable than the initial one (we made the arctangent function disappear). The constants $K,\log 2$ and 
$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32} \tag{7}$$
(see here for the last identity) should simply appear by integration by parts.

With a suitable change of variable and differentiation under the integral sign, we may probably also exploit the integral remainder term in the second Binet's formula for $\log\Gamma$.
A: 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,
A: Hint:
set $x=e^{-y}$ we have
\begin{align}
  & \int_{0}^{1}{\frac{{{\tan }^{-1}}x\,\,\ln x}{1+x}}\,dx=\int_{0}^{\infty }{\,\frac{-y\,{{e}^{-y}}{{\tan }^{-1}}({{e}^{-y}})\,}{1+{{e}^{-y}}}}\,dy \\ 
\\
 & {-{e}^{-y}}{{\tan }^{-1}}({{e}^{-y}})=-{e}^{-y}\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n+1}}}{2n-1}{{e}^{-(2n-1)y}}}=\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{2n-1}{{e}^{-2n\,y}}} \\ 
\\
 & \frac{1}{1+{{e}^{-y}}}=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}{{e}^{-ny}}} \\ 
\end{align}
A: Different approach:
start with applying integration by parts
$$I=\int_0^1\frac{\tan^{-1}(x)\ln(x)}{1+x}dx\\=\left|(\operatorname{Li}_2(-x)+\ln(x)\ln(1+x))\tan^{-1}(x)\right|_0^1-\int_0^1\frac{\operatorname{Li}_2(-x)+\ln(x)\ln(1+x)}{1+x^2}dx$$
$$=-\frac{\pi^3}{48}-\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx-\color{blue}{\int_0^1\frac{\ln(x)\ln(1+x)}{1+x^2}dx}\tag1$$

From $$\operatorname{Li}_2(x)=-\int_0^1\frac{x\ln(y)}{1-xy}dy$$
it follows that
$$\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx=\int_0^1\frac1{1+x^2}\left(\int_0^1\frac{x\ln(y)}{1+xy}dy\right)dx$$
$$=\int_0^1\ln(y)\left(\int_0^1\frac{x}{(1+x^2)(1+yx)}dx\right)dy$$
$$=\int_0^1\ln(y)\left(\frac{\pi}{4}\frac{y}{1+y^2}-\frac{\ln(1+y)}{1+y^2}+\frac{\ln(2)}{2(1+y^2)}\right)dy$$
$$=-\frac{\pi^3}{192}-\color{blue}{\int_0^1\frac{\ln(y)\ln(1+y)}{1+y^2}dy}-\frac12\ln(2)\ G\tag2$$
By plugging $(2)$ in $(1)$, the blue integral magically cancels out and we get $I=\frac12G\ln2-\frac{\pi^3}{64}$.
