<s>[INST] How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$ 
How to evaluate $$\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$$

I tried to integrate by parts, but no way so far, help me, thanks.
 [/INST] different approach to evaluate $\displaystyle\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx$ :
from here , we have $\displaystyle\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi^3}{16}\tag{1}$
and from here , we have $\displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0\tag{2}$
by combining $(1)$ and $(2)$, we obtain that $\displaystyle\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=3\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx+\frac{3\pi^3}{32}\tag{3}$
we have
\begin{align}
\int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx&=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1 x^{2n}\ln(1-x)\ dx\\
&=-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}=-\text{Im}\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\\
\end{align}
and using the generating function with $\ x=i$ $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
we get $\ \displaystyle\int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi}{16}\ln^22+\frac12G\ln2+\text{Im}\operatorname{Li}_3(1-i)\tag{4}$
plugging $(4)$ in $(3)$, we get $$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2+3\text{Im}\operatorname{Li}_3(1-i)$$
</s>