<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] 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.
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