<s>[INST] Compute $\iint_D\frac{xy}{(1+y^2)^2}\,\mathrm{d}x\,\mathrm{d}y$ where $D=\{(x,y):x\geq 0,\ y\geq 0,\ x^2+y^2 \leq 1\}$ 
Compute $$I=\iint_D\frac{xy}{(1+y^2)^2}\,dx\,dy,$$ where $D = \{(x,y):x\geq
 0, \ y\geq 0,\ x^2+y^2 \leq 1\}.$

In the $xy-$plane, this is just a quarter circle disk in the first quadrant. So choosing to integrate first w.r.t $x$ then $y$ I get that
$$I=\int_{0}^{1}\left(\int_0^{\sqrt{1-y^2}}\frac{xy}{(1+y^2)^2}dx\right)dy=\int_0^1\frac{y}{(1+y^2)^2}\left(\int_0^{\sqrt{1-y^2}}x \ dx\right)dy \\ = -\frac{1}{2}\int_0^1\frac{y^3-y}{(y^2+1)^2}dy.$$
Questions:


*

*This seems to work using partial fractions, I checked it with software. Howver I'm not sure how I should do it. Is this the correct ansatz: $$\frac{y^3-y}{(y^2+1)^2}=\frac{Ay+B}{y^2+1}+\frac{Cy+D}{(y^2+1)^2} \ ??$$

*Is there any other nicer way to solve this problem by making the integration simpler? Polar coordinates or something?

 [/INST] Try this
\begin{eqnarray}
I&=&\int_{0}^{1}\left(\int_0^{\sqrt{1-x^2}}\frac{xy}{(1+y^2)^2}dy\right)dx\\&=&\int_0^1x\left(\int_0^{\sqrt{1-x^2}}\frac{y}{(1+y^2)^2} \ dy\right)dx \\ &=& \int_0^1x\left(-\frac{1}{2(1+y^2)^2} \right)\Large|_{0}^{\sqrt{1-x^2}} dx\\
&=& \int_0^1x\left(-\frac{1}{2(2-x^2)^2}+\frac{1}{2} \right)dx\\
&=& \int_0^1\left(-\frac{x}{2(2-x^2)^2}+\frac{x}{2} \right)dx\\
\end{eqnarray}
</s>