$\int_0^\infty\frac{dx}{(1+x^2)(1+9x^2)}$

Use the rational function \[ F(z)={1\over(1+z^2)(1+9z^2)} \] and integrate over the upper semicircle of radius \(R>1\). The arc integral tends to \(0\) because \(F(z)=O(|z|^{-4})\).

The poles in the upper half-plane are \(z=i\) and \(z=i/3\). Their residues are \[ \Res_{z=i}F ={1\over(2i)(1-9)}={i\over16}, \] and \[ \Res_{z=i/3}F ={1\over(1-1/9)(6i)}=-{3i\over16}. \] Thus \[ \int_{-\infty}^{\infty}{dx\over(1+x^2)(1+9x^2)} =2\pi i\left({i\over16}-{3i\over16}\right) ={\pi\over4}. \] The integrand is even, so \[ \boxed{ \int_0^\infty {dx\over(1+x^2)(1+9x^2)}={\pi\over8}.} \]