\[ \int_\RR {1 \over (1+x^2)^2}\dx = {\pi \over 2} .\]
Solution
Let \[ f(z)=\frac{1}{(1+z^2)^2}. \] Integrate over the positively oriented upper semicircle of radius \(R>1\). The only enclosed pole is the double pole at \(z=i\), with residue \[ \Res_{z=i}f(z) =\left.\dd{}{z}\frac{1}{(z+i)^2}\right|_{z=i} =-\frac{2}{(2i)^3} =\frac{1}{4i}. \] Hence \[ \int_{-R}^R f(x)\,dx+\int_{C_R}f(z)\,dz =2\pi i\Res_{z=i}f(z) =\frac{\pi}{2}. \] On \(C_R\), \[ |1+z^2|\ge R^2-1, \] so the ML estimate gives \[ \abs{\int_{C_R}f(z)\,dz} \le \pi R\frac{1}{(R^2-1)^2}\to0. \] Letting \(R\to\infty\) yields \[ \int_{\RR}\frac{dx}{(1+x^2)^2}=\frac{\pi}{2}. \]