<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] Through the dilogarithm/trilogarithm machinery it can be shown that
$$ \int_{0}^{1}\frac{\log(1+i x)\log(1+x)}{x}\,dx=\\\frac{\pi K}{2}-\frac{9i\pi^3}{64}+3iK\log(2)-\frac{3\pi i}{16}\log^2(2)+\frac{5\pi^2}{32}\log(2)-\frac{\log^3(2)}{8}-\frac{69}{16}\zeta(3)+6\,\text{Li}_3\left(\tfrac{1+i}{2}\right) $$
$$ \int_{0}^{1}\frac{\log^2(1+i x)}{x}\,dx=\\
-\frac{\pi K}{2}-\frac{3i\pi^3}{64}+iK\log(2)-\frac{\pi i}{16}\log^2(2)+\frac{5\pi^2}{96}\log(2)-\frac{\log^3(2)}{24}-\frac{3}{16}\zeta(3)+2\,\text{Li}_3\left(\tfrac{1+i}{2}\right) $$
$$ \int_{0}^{1}\frac{\log(1+ix)\log(1-ix)}{x}\,dx= \frac{\pi K}{2}-\frac{27}{32}\zeta(3)$$
hence the claim follows by $\arctan x=\text{Im}\,\log(1+ix)$ and $\log(1+x^2)=\log(1+ix)+\log(1-ix)$.
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