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.
 A: Solution 1. Let $I$ denote the double integral. Then
\begin{align*}
I
&= \int_0^1 \int_0^\infty \frac{1}{(1+x^2)(1+x^2 y^2)} \, \mathrm{d}x\,\mathrm{d}y \\
&= \int_0^1 \int_0^\infty \frac{1}{1-y^2} \left( \frac{1}{1+x^2} - \frac{y^2}{1+x^2y^2} \right) \, \mathrm{d}x\,\mathrm{d}y \\
&= \int_{0}^{1} \frac{1}{1-y^2} \left( \frac{\pi}{2} - \frac{\pi y}{2} \right) \, \mathrm{d}y \\
&= \frac{\pi}{2} \int_0^1 \frac{1}{1+y} \, \mathrm{d}y \\
&= \frac{\pi}{2} \log 2.
\end{align*}
Solution 2.
\begin{align*}
I
= \int_0^\infty \int_0^1 \frac{1}{(1+x^2)(1+x^2 y^2)} \, \mathrm{d}y \, \mathrm{d}x
= \int_0^\infty \frac{\arctan x}{(1+x^2)x} \, \mathrm{d}x.
\end{align*}
Substituting $x = \tan\theta$, then
\begin{align*}
\require{cancel}
I
&= \int_0^{\frac{\pi}{2}} \frac{\theta}{\tan\theta} \, \mathrm{d}\theta
= \cancel{\left[ \theta \log \sin\theta \right]_0^{\frac{\pi}{2}}} - \int_0^{\frac{\pi}{2}} \log \sin\theta \, \mathrm{d}\theta.
\end{align*}
Regarding the last integral, we observe that the following holds:
$$ I
= -\int_0^{\frac{\pi}{2}} \log \sin\theta \, \mathrm{d}\theta
= -\int_0^{\frac{\pi}{2}} \log \cos\theta \, \mathrm{d}\theta
= -\int_0^{\frac{\pi}{2}} \log \sin(2\theta) \, \mathrm{d}\theta. $$
Using this, we get
$$ 2I
= -\int_0^{\frac{\pi}{2}} \log (\sin\theta\cos\theta) \, \mathrm{d}\theta
= -\int_0^{\frac{\pi}{2}} \log \left(\frac{\sin2\theta}{2}\right) \, \mathrm{d}\theta
= I + \frac{\pi}{2}\log 2. $$
Therefore
$$ I = \frac{\pi}{2}\log 2. $$
A: 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
$$
\begin{eqnarray}
\iint_{A=[0,\infty[\times[0,1]}\frac{dx\,dy}{(1+x^2)(1+x^2y^2)}&=&\int_0^1\frac{1}{1-y^2}\left[\int_0^\infty\left(\frac{1}{1-x^2}-\frac{y^2}{1+x^2y^2}\right)dx\right]dy \\
&=&\int_0^1\frac{1}{1-y^2}\left[\arctan x|_0^\infty-y\arctan (xy)|_0^\infty\right]dy \\
&=&\int_0^1\frac{1}{1-y^2}\left[\frac{\pi}{2}-y\frac{\pi}{2}\right]dy \\
&=&\frac{\pi}{2}\int_0^1\frac{dy}{1+y}=\left.\frac{\pi}{2}\ln|1+y|\right|_0^1 \\
&=&\frac{\pi}{2}\ln 2
\end{eqnarray}
$$
A: \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 $$ \int_0^{\pi/2} \frac{w}{\tan w} \, dw = \frac \pi 2 \log 2. $$
