<s>[INST] Integrate $\iint_D \frac1{(x^2+y^2)^2 }dx dy$ over a region bounded by four circles 

I'm thinking of doing the substitution $x = \frac{u}{u^2 + v^2}, y = \frac{v}{u^2 + v^2},$ but I'm not sure how to exactly compute the range of values $u$ and $v$ take. Clearly, $u=\frac{x}{x^2+y^2}$ and $v = \frac{y}{x^2 + y^2}.$ The Jacobian of the result is $-\frac{1}{u^2 + v^2}$, and so the resulting integral equals $\iint_{D'} dudv,$ where $D'$ is the region of possible values for the pairs $(u,v)$. In the circle $x^2 + y^2 - 2x= 0$, for instance, we have the point $(1,0)$, which corresponds to $u=1, v=0$. Also, we have the point $(2,0)$, corresponding to $u= \frac{1}2, v = 0$. I can't seem to generalize what values u and v can take on.

 [/INST] In polar coordinates, the enclosed region can be partitioned into three angular spans, with break points corresponding to the intercepts of the circles, at $\tan^{-1}\frac13, \tan^{-1}\frac23, \frac\pi4$ and $\tan^{-1}2$. Then
\begin{align}
&\iint_D \frac1{(x^2+y^2)^2 }dx dy\\
=&  \int_{\tan^{-1}\frac13}^{\tan^{-1}\frac23}\int_{2\cos\theta}^{6\sin\theta}
 + \int_{\tan^{-1}\frac23}^{\frac\pi4}\int_{2\cos\theta}^{4\cos\theta}
 + \int_{\frac\pi4}^{\tan^{-1}2}\int_{2\sin\theta}^{4\cos\theta}\frac1{r^3} dr d\theta\\
=&\>\frac1{48}+ \frac1{32} +\frac1{32}=\frac1{12}
\end{align}

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