Finding $\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$ with Fubini's theorem I have to solve the following double integral
$$\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$$
with $A= \left[0,+\infty\right[ \times [0,1].$ So far I've tried to solve it integrating w.r.t. $y$ first.
$$\iint_0^1\frac{dy\,dx}{(1+x^2)(1+x^2 y^2)} = \int_0^\infty\frac{1}{1+x^2}\int_0^1\frac{dy}{1+x^2 y^2} \, dx. $$
I've solved the internal integral by substitution, remembering that $\int\frac{du}{1+u^2}=\arctan u$
Substitution:
$$x^2 y^2= u^2 \to y=\frac{1}{x}u \to dy=\frac{1}{x}du.$$
$$y=0 \to u=0, \qquad y=1 →u=x$$
So:
\begin{align}
& \int_0^∞\frac{1}{1+x^2} \left( \int_0^x\frac{1}{x}\frac{1}{1+u^2}\,du \right) \, dx \\[8pt]
= {} & \int_0^\infty\frac{1}{x}\frac{1}{1+x^2}[\arctan u] \, dx \\[8pt]
= {} & \int_0^\infty\frac{\arctan x}{x(1+x^2)} \, dx.
\end{align}
Now I have to solve this last integral with the Fubini's theorem but I don't know how to do it.

\begin{align} & \iint\limits_{[0,+\infty) \times [0,1]} \frac{d(x,y)}{(1+x^2)(1+x^2 y^2)} \\[8pt] = {} & \int_0^1 \left( \int_0^\infty \frac{dx}{(1+x^2)(1+x^2y^2)} \right) \, dy \end{align}
So we need partial fractions:
\begin{align}
& \frac 1 {(1+x^2)(1+x^2 y^2)} = \frac {Ax+B} {1+x^2} + \frac{Cx+D}{1+x^2y^2} \\[10pt]
1 & = (Ax+B)(1+x^2y^2) + (Cx+D)(1+x^2) \\[8pt]
& = (Ay^2+C)x^3 + (By^2+D)x^2 + (A+C)x + (B+D) \\[8pt]
\text{So } & Ay^2 + C=0 \\
& By^2+D=0 \\
& A+C=0 \\
& B+D=1 \\[8pt]
\text{and so } & A=0, \quad C=0, \quad B=1/(1-y^2), \quad D= y^2/(y^2-1). \\[8pt]
& \frac 1 {(1+x^2)(1+x^2 y^2)} = \frac {1/(1-y^2)} {1+x^2} + \frac{y^2/(y^2-1)}{1+x^2y^2}
\end{align}
And so the first integral:
\begin{align}
& \int_0^\infty \frac{dx}{(1+x^2)(1+x^2y^2)} \\[8pt]
= {} & \int_0^\infty \left( \frac{1/(1-y^2)} {1+x^2} + \frac{y^2/(y^2-1)}{1+x^2y^2} \right) \, dx \\[8pt]
= {} & \frac \pi {2(1+y)}
\end{align}
and so on.
Corollary: In view of one of my comments under the question, we conclude that $$ \int_0^{\pi/2} \frac{w}{\tan w} \, dw = \frac \pi 2 \log 2. $$
