Evaluating $\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$ I am trying to evaluate this integral 
$$\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$$
with $\mathcal{P}$ the principal value and $a,b>0$.
I already know the answer to be $$ - \frac{a\pi}{2(1 + a^2b^2)}$$
after fiddling with Mathematica putting numbers for $a$ and $b$.
The poles of this function are at $x=\pm b, \pm \frac{i}{a^2}$ so I cannot find a proper contour.
Any help in getting the answer will be appreciated!

Thanks to both of you for the help!
So indeed, this contour is a good one to evaluate this integral.
The residues at $b$ and $-b$ just cancel each other.
The residue at $i/a$ gives $$ - \frac{a}{2i\left( a^2 b^2 +1 \right)}$$
Thus we have
$$ 2\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)} = 2\pi i \left(- \frac{a}{2i\left( a^2 b^2 +1 \right)} \right) =  \frac{-\pi a}{1 + a^2b^2}$$
$$ \mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)} = -\frac{\pi}{2}\frac{a}{1 + a^2b^2}$$
