calculate improper integral over $\mathbb{R}^2$

Question

I was asked to prove that the improper integral $$\int_{\mathbb{R}^2}{\dfrac{1}{\left(1+4x^2+9y^2\right)^3}}dxdy$$ exists and calculate its value.
I immediately thought about change of variable using polar coordinates, such that: $$\varphi:\mathbb{R}^2\rightarrow\mathbb{R}^2$$ $$\varphi(r,\theta)=(r\cos\theta,r\sin\theta)$$ $$\int_{\mathbb{R}^2}{\dfrac{1}{\left(1+4x^2+9y^2\right)^3}}dxdy = \lim_{R\rightarrow\infty}{\int_{B(0,R)}{\dfrac{1}{\left(1+4x^2+9y^2\right)^3}}dxdy}$$ where $B(0,R)$ is the unit sphere in $\mathbb{R}^2$ around $0$ with radius $R>0$.
we know that: $$\varphi([0,R]\times[0,2\pi])=B(0,R)$$ so from change of variable we get that: $$\int_{B(0,R)}{\dfrac{1}{\left(1+4x^2+9y^2\right)^3}}dxdy=\int_{[0,R]\times[0,2\pi]}{\dfrac{r}{\left(1+4{(r\cos\theta)}^2+9{(r\sin\theta)}^2\right)^3}} \cdot d\theta dr$$when $r$ came in because $|\det (\varphi^{'})|=r$
from here on I couldn't simplify the expression I got as $4$ and $9$ does not have a common divisor so I'm stuck with $\sin$ and $\cos$.
Would glad to get some suggestions on how to keep going.

Answer 1

First, apply the transformation $t=2x$ and $u=3y$ to get $$6\int_{\mathbb R^2}\frac1{(1+t^2+u^2)^3}dtdu.$$ Now, use polar coordinates: the integral over $\theta$ is easy, and you're just left with a constant times $$\int_0^\infty \frac{rdr}{(1+r^2)^3}$$ which is much simpler to evaluate.