After trying many methods but fail, I try the following substitution.
Letting $t=\frac{1-x}{1+x}$ preserves the interval and transforms the integral into $$I=\int_0^1 \arctan x \ln \left(\frac{1-x}{1+x}\right) d x = 2 \int_0^1 \arctan \left(\frac{1-t}{1+t}\right) \frac{\ln t}{(1+t)^2} d t$$ Noting that $$\tan \left(\frac{\pi}{4}-\arctan t\right)=\frac{1-t}{1+t} \Leftrightarrow \frac{\pi}{4}-\arctan t=\arctan \left(\frac{1-t}{1+t}\right) $$ $$ I=\frac{\pi}{2} \underbrace{\int_0^1 \frac{\ln t}{(1+t)^2} d t }_K-2 \underbrace{\int_0^1\frac{\arctan t}{(1+t)^2} \ln t d t}_L $$
Using integration by parts, we have $$ \begin{aligned} \int \frac{\ln t}{(1+t)^2} d t & =-\int \ln t d\left(\frac{1}{1+t}\right)=-\frac{\ln t}{1+t}+\int \frac{d t}{t(1+t)} \\ & =-\frac{\ln t}{1+t}+\ln t-\ln (1+t)=\frac{t \ln t}{1+t}-\ln (1+t) \cdots (*) \\\Rightarrow \quad K&=-\ln 2 \end{aligned} $$
Using (*), we can tackle the integral $L$ by integration by parts again, $$ \begin{aligned} L= & \underbrace{ {\left[\arctan t\left(\frac{t \ln t}{1+t}-\ln (1+t)\right)\right]_0^1 }}_{=-\frac{\pi}{4}\ln 2 } - \underbrace{ \int_0^1 \frac{t \ln t}{\left(1+t^2\right)(1+t)} d t}_{M} + \underbrace{ \int_0^1 \frac{\ln (1+t)}{1+t^2} d t}_{N} \end{aligned} $$
Using power series and partial fractions, $$ \begin{aligned} M&=-\frac{1}{2} \int_0^1 \frac{\ln t}{1+t} d t+\frac{1}{2} \int_0^1 \frac{t \ln t}{1+t^2} d t+\frac{1}{2} \int_0^1 \frac{\ln t}{1+t^2} dt \\&=-\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^k \ln t d t+\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{2k +1} \ln tdt +\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{2 k} \ln t d t\\&= -\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}+\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+2)^2}+\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^2}\\&= -\frac{3}{8}\left(1-\frac{1}{2}\right) \zeta(2)+\frac{1}{2} G\\&=-\frac{\pi^2}{32}+\frac{G}{2}\end{aligned} $$
Letting $t=\tan \theta$ transforms $$ N=\int_0^{\frac{\pi}{4}} \ln (1+\tan \theta) d \theta \stackrel{x\mapsto\frac{\pi}{4}-x}{=} \int_0^{\frac{\pi}{4}} \ln \left(\frac{2}{1+\tan \theta}\right) d \theta= \frac{\pi}{8} \ln 2 $$
Grouping them together yields $$I=-\frac{\pi}{4} \ln 2+\frac{\pi^2}{16}-G$$
Is there a simple way to evaluate $\int_0^1 \arctan x \ln \left(\frac{1-x}{1+x}\right) d x?$
Your comments and alternatives are highly appreciated.
