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.

Following Michael Hardy's suggestion, I will use partial fractions. We will have
$$\frac{1}{(1+x^2)(1+x^2y^2)}=\frac{1}{1-y^2}\left(\frac{1}{1-x^2}-\frac{y^2}{1+x^2y^2}\right)$$
So
$$
\begin{eqnarray}
\iint_{A=[0,\infty[\times[0,1]}\frac{dx\,dy}{(1+x^2)(1+x^2y^2)}&=&\int_0^1\frac{1}{1-y^2}\left[\int_0^\infty\left(\frac{1}{1-x^2}-\frac{y^2}{1+x^2y^2}\right)dx\right]dy \\
&=&\int_0^1\frac{1}{1-y^2}\left[\arctan x|_0^\infty-y\arctan (xy)|_0^\infty\right]dy \\
&=&\int_0^1\frac{1}{1-y^2}\left[\frac{\pi}{2}-y\frac{\pi}{2}\right]dy \\
&=&\frac{\pi}{2}\int_0^1\frac{dy}{1+y}=\left.\frac{\pi}{2}\ln|1+y|\right|_0^1 \\
&=&\frac{\pi}{2}\ln 2
\end{eqnarray}
$$
