Nice Integral : $$\int^1_0 \arctan\left({\frac{1-x}{1+x}}\right)\arctan(x)\frac{1}{1-x}dx$$
Let : $x:=\frac{1-x}{1+x}$ $$I=\int^1_0 \arctan\left({\frac{1-x}{1+x}}\right)\arctan(x)\frac{1}{x(1+x)}dx$$ I need help with this integration because it seems difficult to me. I appreciate your interest.
\begin{align} I=&\int^1_0 \frac{\tan^{-1}{\frac{1-x}{1+x}}\tan^{-1}x}{1-x}dx\\ \overset{ibp}=& \int^1_0 \frac{\ln(1-x)\left( \tan^{-1}{\frac{1-x}{1+x}}-\tan^{-1}x\right)}{1+x^2}\overset{x\to \frac{1-x}{1+x}}{ dx}\\ =& \int^1_0 \frac{\ln\frac{2x}{1+x}\left(2 \tan^{-1}x -\frac\pi4\right)}{1+x^2}{ dx}\\ =& \int^1_0 \frac{2\ln\frac{4x}{1+x^2} \tan^{-1}x }{1+x^2}dx +\int^1_0 \frac{2\ln\frac{1+x^2}{2(1+x)}\tan^{-1}x }{1+x^2}-\frac{\frac\pi4\ln\frac{2x}{1+x}}{1+x^2}\ dx \end{align} Note that the 2nd integral vanishes as seen below \begin{align} &\int^1_0 \underset{x\to \frac{1-x}{1+x}}{\frac{2\ln\frac{1+x^2}{2(1+x)}\tan^{-1}x }{1+x^2}}{ dx} -\int^1_0 \frac{\frac\pi4 \ln\frac{2x}{1+x}}{1+x^2}dx= \int^1_0 \frac{\frac\pi4 \ln\frac{1+x^2}{4x}}{1+x^2}dx=0 \end{align} Thus $$I= \int^1_0 \frac{2\ln\frac{4x}{1+x^2} \tan^{-1}x }{1+x^2}{dx} \overset{x=\tan \frac t2} =\int_0^{\frac\pi2}\frac t2\ln(2\sin t)dt= \frac{7}{32}\zeta(3) $$ See here for the last step.
Note that $\tan^{-1}(\frac{1-x}{1+x})=\frac{\pi}{4}-\tan^{-1}(x)$. With partial fraction $\frac{1}{x(1+x)}=\frac{1}{x}-\frac{1}{1+x}$
We have $$I=\int_0^1\tan^{-1}\left(\frac{1-x}{1+x}\right)\tan^{-1}(x)\frac{dx}{x(1+x)}$$
$$=\frac{\pi}{4}\int_0^1\frac{\text{arctan} x}{x}\, dx-\int_0^1 \frac{\text{arctan}^2 x}{x}\, dx-\frac{\pi}{4}\int_0^1\frac{\text{arctan} x}{1+x}\, dx+\int_0^1\frac{\text{arctan}^2 x}{1+x}\, dx$$
The values of the first few integrals could be calculated rather easily
$$\int_0^1\frac{\text{arctan} x}{x}\, dx=G$$
$$\int_0^1 \frac{\text{arctan}^2 x}{x}\, dx=\frac{\pi G}{2}-\frac{7}{8}\zeta(3)$$
$$\int_0^1\frac{\text{arctan} x}{1+x}\, dx=\frac{\pi}{8}\log(2)$$
Our main challenge is the last integral. Integration by parts gives
$$\int_0^1\frac{\text{arctan}^2 x}{1+x}\, dx=\frac{\pi^2}{16}\log(2)-{\underbrace{\int_0^1 \frac{2 \text{arctan} x}{1+x^2}\log(1+x)\, dx}_{J}}$$
$$J=\int_0^1\int_0^1\frac{2x \arctan x}{(1+x^2)(1+yx)}\, dy\, dx$$ Note the partial fraction
$$\frac{x}{(1+x^2)(1+xy)}=\frac{1}{1+y^2}\left(\frac{x}{1+x^2}+\frac{y}{1+x^2}-\frac{y}{1+xy}\right)$$
$$\begin{split}\implies J=& \int_0^1\frac{dy}{1+y^2}\int_0^1 \frac{2x}{1+x^2}\text{arctan}(x)\, dx+\int_0^1 \frac{2y}{1+y^2}\, dy\int_0^1\frac{\text{arctan} x}{1+x^2}\, dx\\ & -\int_0^1 \int_0^1 \frac{2y\arctan(x)}{(1+y^2)(1+xy)}\, dx\, dy\end{split}$$
The integral $$\int_0^1 \frac{2x}{1+x^2}\text{arctan}(x)\, dx=G-\frac{\pi}{4}\log(2)$$ can be proved using integration by parts and Fourier series of $\log(\cos x)$
$$J=\frac{\pi}{4}G-\frac{\pi^2}{16}\log(2)+\frac{\pi^2}{32}\log(2)-\int_0^1\frac{2y}{1+y^2}\int_0^1\frac{\text{arctan} x}{1+xy}\, dx\, dy$$ $$=\frac{\pi}{4}G-\frac{\pi^2}{32}\log(2)-\int_0^1\frac{2y}{1+y^2}\left[\frac{\pi \log(1+y)}{4y}-\int_0^1 \frac{\log(1+xy)}{y(1+x^2)}\, dx\,\right] dy$$ $$=\frac{\pi}{4}G-\frac{\pi^2}{32}\log(2)-\frac{\pi}{2}\int_0^1\frac{\log(1+y)}{1+y^2}\, dy+2\int_0^1\int_0^1 \frac{\log(1+xy)}{(1+x^2)(1+y^2)}\, dx\, dy$$ Note that $$\int_0^1\frac{\log(1+y)}{1+y^2}\, dy=\frac{\pi}{8}\log(2)$$ Can be proved by substituting $x\mapsto \frac{1-x}{1+x}$.
Grouping the results gives
$$J=\frac{\pi}{4}G-\frac{3\pi^2}{32}\log(2)+2L$$ where $\displaystyle L=\int_0^1\int_0^1 \frac{\log(1+xy)}{(1+x^2)(1+y^2)}\, dx\, dy$
Which writing $I$ in terms of $L$ gives the equation
$$I=-\frac{\pi}{2}G+\frac{7}{8}\zeta(3)+\frac{\pi^2}{8}\log(2)-2L$$
$L$ looks symmetric and reasonable to compute, I will leave the details to OP.
$$\begin{align*} I &= \int_0^1 \arctan \frac{1-x}{1+x} \cdot \arctan x \cdot \frac{dx}{1-x} \\ &= \int_0^1 \frac{\left(\frac\pi4-\arctan x\right) \arctan x}{1-x} \, dx \\ I_1 &= \int_0^\tfrac\pi4 \frac{x \left(\frac\pi4-x\right)}{\cos x(\cos x-\sin x)} \, dx & x\to\tan x \\ I_2 &= \int_0^\tfrac\pi4 \frac{x \left(\frac\pi4-x\right)}{\sin x(\cos x+\sin x)} \, dx & x\to\frac\pi4-x \\ I = \frac{I_1+I_2}2 &= \int_0^\tfrac\pi4 \frac{2x \left(\frac\pi4-x\right)}{\sin(4x)} \, dx \\ &= \frac1{32} \int_0^\pi \frac{x(\pi-x)}{\sin x} \, dx & x\to\frac x4 \\ &= \frac1{16} \int_0^\tfrac\pi2 \frac{x(\pi-x)}{\sin x} \, dx & \rm symmetry \\ &= \frac1{16} \left(2\pi G - \left(2\pi G - \frac72 \zeta(3)\right)\right) = \boxed{\frac7{32}\zeta(3)} \end{align*}$$
where $G$ denotes Catalan's constant. See also the integral of $x^2\csc x$.
