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.

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 $I=\frac12G\ln2-\frac{\pi^3}{64}$.
