$$ \int_{-\infty}^{\infty} \frac{dz}{(z^2+a^2)^2}$$ Now this has the poles at $$(z^2+a^2) = 0 $$
if we consider the contour from -R to +R extending from $-\infty$ to $+\infty$ i get the roots $z = ae^{i\pi/2},ae^{i3\pi/2}$ that lie in the contour Now how to find the residues of these poles ??
Bear in mind that for $a\in\Bbb R\setminus\{0\}$ the pole $ai$ is second-order, with residue $$\lim_{z\to ai}\frac{d}{dz}\frac{1}{(z\pm ai)^2}=\frac{-2}{(2ai)^3}=-\frac{i}{4a^3}.$$Multiplying by $2\pi i$ gives the integral's value, $\frac{\pi}{2a^3}$. (The pole at $-ai$ isn't enclosed in the contour, so doesn't contribute to the integral.)
It always helps to double-check with other methods when you're new to this:$$z=a\tan t\implies\int_{\Bbb R}\frac{dz}{(z^2+a^2)^2}=\frac{1}{a^3}\int_{-\pi/2}^{\pi/2}\cos^2 tdt=\frac{\pi}{2a^3}.$$