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.
 A: 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 $\log(1+x^2)=\log(1+ix)+\log(1-ix)$.
A: @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 
$$\int_0^1\tan^{-1}(x)\ln\left(\frac{(1+x^2)^3}{(1+x)^2}\right)dx=0$$
A: A solution by Cornel Ioan Valean. The problem is similar to the problem AMM $12054$. Using the well-known result in $4.535.1$ from  Table of Integrals, Series and Products by I.S. Gradshteyn and I.M. Ryzhik:
$$\int_0^1 \frac{\arctan(y x)}{1+y^2x}\textrm{d}x=\frac{1}{2y^2}\arctan(y)\log(1+y^2)$$
We have:
$$\frac{1}{2}\int_0^1\frac{\arctan(y)\log(1+y^2)}{y}dy=\int_0^1\left(\int_0^1 \frac{y\arctan(y x)}{1+y^2x}\textrm{d}x\right)\textrm{d}y$$
$$\overset{yx=t}{=}\int_0^1\left(\int_0^y \frac{\arctan(t)}{1+y t}\textrm{d}t\right)\textrm{d}y=\int_0^1\left(\int_t^1 \frac{\arctan(t)}{1+y t}\textrm{d}y\right)\textrm{d}t$$
$$=\int_0^1\frac{\arctan(t)\log\left(\frac{1+t}{1+t^2}\right)}{t} \textrm{d}t\overset{t=y}=\int_0^1\frac{\arctan(y)\log\left(\frac{1+y}{1+y^2}\right)}{y} \textrm{d}y$$
And the result is proved.
