I know you can change to polar coordinates but then you still have to integrate $\sqrt{1+r^2}$ which is still non-trivial.
I remember there being some trigonometric substitution, possibly hyperbolic (not sure!) that made life much easier.
Anyone has any idea?
No, you don't have to integrate $\sqrt{1+r^2}$. What you have to integrate is $r\sqrt{1+r^2}$, which is quite simple, where the first $r$ is the area element in polar coordinates.