Evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$

Question

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.

Answer 1

From here we have that $$\frac12 \int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac13 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$ $$\Rightarrow I=\int_{0}^{1}\frac{\arctan x}{x} \ln{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)} dx=\frac23 \int_{0}^{1}\frac{\arctan x \ln(1+x)}{x} dx$$ I have encountered this integral too last year and asked it on AoPS, you can take a look at Knas solution from there, giving: $$I=\begin{align}2\Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\text{G}\ln 2-\frac{3}{64}\pi^3-\frac{1}{16}\pi\ln^2 2\end{align}$$