Compute $\iint_D\frac{xy}{(1+y^2)^2}\,\mathrm{d}x\,\mathrm{d}y$ where $D=\{(x,y):x\geq 0,\ y\geq 0,\ x^2+y^2 \leq 1\}$

Question

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:

1. 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: $$\frac{y^3-y}{(y^2+1)^2}=\frac{Ay+B}{y^2+1}+\frac{Cy+D}{(y^2+1)^2} \ ??$$
2. Is there any other nicer way to solve this problem by making the integration simpler? Polar coordinates or something?

Answer 1

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}

Answer 2

By switching to polar coordinates the given integral equals

$$ \int_{0}^{1}\int_{0}^{\pi/2}\frac{\rho^3\sin(\theta)\cos(\theta)}{(1+\rho^2 \sin^2(\theta))^2}\,d\theta\,d\rho=\int_{0}^{1}\frac{\rho^3\,d\rho}{2+2\rho^2}=\color{red}{\frac{1-\log 2}{4}}. $$