this is an even problem in my textbook. So this is one of the few places I can check my answer.
$$\int _{-1}^1\int _{-\sqrt{1\:-\:x^2}}^{\sqrt{1-x^2}}\:\frac{2}{\left(1\:+\:x^2\:+\:y^2\right)^2}\:dy\:dx$$
this is what I converted it to
$$\int _0^{2\pi }\int _0^1\:\frac{2r}{\left(1\:+\:r^2\right)^2}\:dr\:d\theta $$
I'm not sure if this is the correct conversion. If it is indeed correct, how should go about integrating it?
On
$$\int _{-1}^1\int _{-\sqrt{1\:-\:x^2}}^{\sqrt{1-x^2}}\:\frac{2}{\left(1\:+\:x^2\:+\:y^2\right)^2}\:dy\:dx$$
convert the integral above to:
$$\int _0^{2\pi }\int _0^1\:\frac{2r}{\left(1\:+\:r^2\right)^2}\:dr\:d\theta $$
use U-sub: $$\:u\:=\:1\:+\:r^2$$
$$\int _0^{2\pi }\int _1^2\:\frac{1}{u^2}\:dud\theta $$
integrate the inner integral and get
$$ \int _0^{2\pi }\:\frac{1}{2}\:d\theta $$
integrate above and get
$$ \frac{1}{2}\left(2\pi \right)\:-\:\frac{1}{2}\left(0\right)\:=\:\pi $$
It is correct. Readily recognizable integrate $ \frac{-1}{1+r^2}.. $