= Solution
Apply the <residue theorem> to $F(z)=(z^2+4)^{-2}$ on the upper semicircle of radius $R>2$, traversed counterclockwise. Its sole enclosed <pole> is the double pole at $2i$, and
$$
\operatorname{Res}_{z=2i}F
=\left.\frac d{dz}(z+2i)^{-2}\right|_{z=2i}
=-\frac2{(4i)^3}=-\frac i{32}.
$$
On the semicircular arc, $|z^2+4|\ge R^2-4$, so the <ML inequality> bounds its contribution by $\pi R/(R^2-4)^2\to0$. The real integral is absolutely convergent, since its integrand is $O(|x|^{-4})$ at infinity. Taking $R\to\infty$ in the contour identity yields
$$
\boxed{\int_{-\infty}^{\infty}\frac{dx}{(x^2+4)^2}=2\pi i\left(-\frac i{32}\right)=\frac\pi{16}.}
$$
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