How to prove $\int_{\mathbb R^2}\frac{dxdy}{(x^2+y^2+1)^{3/2}}=2\pi$ I want to use polar coordinates to prove this:
$$\int_{\mathbb R^2}\frac{dxdy}{(x^2+y^2+1)^{3/2}}=2\pi$$
I'm thinking to make the following substitution:
$x^2+y^2=r^2, dx=-r\sin\theta d\theta, dy=r\cos\theta d\theta$
Am I in the right way? I need help how to proceed.
 A: Your way to reason is correct.
Apply a polar coordinates transformation as you wrote, to obtain:
$$\text{d}x\text{d}y \to \text{Jacobian} =  r\ \text{d}r\ \text{d}\theta$$
$$x^2 + y^2 = r^2$$
Hence the integral becomes
$$\int_0^{+\infty}\int_0^{2\pi} \frac{r\ \text{d}r\ \text{d}\theta}{(r^2+1)^{3/2}}$$
The integral over $\theta$ is trivial and you get immediately a factor of $2\pi$.
It's not difficult to show thence that
$$\int_0^{+\infty} \frac{r}{(r^2+1)^{3/2}}\ \text{d}r = 1$$
once the final result is 
$$2\pi$$
Write me if you need details to solve the latter integral!
How to solve the integral
With a simple substitution:
$$k = r^2+1$$
$$\text{d}k = 2r\ \text{d}r$$
hence
$$\frac{1}{2}\int\frac{\text{d}k}{k^{3/2}}$$
You can easily proceed by yourself.
A: There is a faster way, that is integrating along "shells":
$$ \iint_{\mathbb{R}^2}\frac{dx\,dy}{(1+x^2+y^2)^{3/2}} = \int_{0}^{+\infty}\frac{2\pi\rho}{(1+\rho^2)^{3/2}}\,d\rho = \left.-\frac{2\pi}{\sqrt{1+\rho^2}}\right|_{0}^{+\infty} = \color{red}{2\pi}$$
since the length of the circle $x^2+y^2=\rho^2$ is just $2\pi\rho$.
