<s>[INST] An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$ 
How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$

I tried substitution $x=\frac{1-t}{1+t}$ and got
$$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\
=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} (\arctan t-\frac\pi4)}{t^2-1}dt$$
I'm able to evaluate $$\int_0^1\frac{\ln \frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
 [/INST] @Kemono Chen proved here
$$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
Divide both sides by $y$ then integrate between $0$ and $1$ we get
$$\color{red}{\frac12\mathcal{I}}=\frac12\int_0^1\frac{\tan^{-1}(y)\ln(1+y^2)}{y}dy=\int_0^1\int_0^y\frac{\ln(1+yx)}{y(1+x^2)}dxdy$$
$$=\int_0^1\frac{1}{1+x^2}\left(\int_x^1\frac{\ln(1+xy)}{y}dy\right)dx=\int_0^1\frac{\operatorname{Li}_2(-x^2)-\operatorname{Li}_2(-x)}{1+x^2}dx$$
$$\overset{IBP}{=}\int_0^1\tan^{-1}(x)\left(\frac{2\ln(1+x^2)}{x}-\frac{\ln(1+x)}{x}\right)dx\\=\color{red}{2\mathcal{I}}-\int_0^1\frac{\tan^{-1}(x)\ln(1+x)}{x}dx$$
which can be written as 
$$\int_0^1\tan^{-1}(x)\ln\left(\frac{(1+x^2)^3}{(1+x)^2}\right)dx=0$$
</s>