This is my initial attempt:
1- $ I = \int_{all \ space} d^2x / (a^2 + r^2)^3 $ where $ d^2x = dx.dy $ and $ r^2=x^2+y^2 $
2- Transformation to polar coord. $ I = \int_{0}^{\infty} \int_{0}^{2 \pi} r.d \theta . dr / (a^2 + r^2)^3 $ . This is an odd function and has also no branch point. I have tried to add a branch point and take a choose a clever contour, but my attempts failed.
For those interested the question is from: An Introduction to Mathematical Methods of Physics (Lorella M. Jones), Section 8 and question 4.
Using the theorem
$$\oint_{\partial D}f(z)\:dz = 2i\iint_D \frac{\partial f}{\partial\bar{z}}dxdy$$
and $x^2+y^2 = z\bar{z}$, consider the following integrals
$$\iint\limits_{r\leq|z|\leq R}\frac{dxdy}{(a^2+z\bar{z})^3} = \frac{1}{4i}\oint_{|z|=r}\frac{dz}{z(a^2+z\bar{z})^2} - \frac{1}{4i}\oint_{|z|=R}\frac{dz}{z(a^2+z\bar{z})^2}$$
$$ = \frac{1}{4}\int_{0}^{2\pi}\frac{d\theta}{(a^2+r^2)^2} - \frac{1}{4}\int_{0}^{2\pi}\frac{d\theta}{(a^2+R^2)^2} = \frac{\pi}{2(a^2+r^2)^2} - \frac{\pi}{2(a^2+R^2)^2}$$
The value of your integral is therefore
$$I = \lim_{r\to0^+}\frac{\pi}{2(a^2+r^2)^2}-\lim_{R\to\infty}\frac{\pi}{2(a^2+R^2)^2} = \boxed{\frac{\pi}{2a^4}}$$