How do I show $\int_{-\infty}^\infty \frac 1{(a^2+s^2)(b^2+s^2)} ds=\frac {\pi}{ab(a+b)}$ using the solution to the following Fourier transform? For a function  $f_a(x)=e^{-a|x|}$ , where $a>0$ I have found that the fourier transform of it is as follows, i know this is correct.
$\def\F{\mathcal F}$
\begin{align*}
  \F(f_a)(s) &=  \sqrt{\frac 2\pi} \frac a{a^2 + s^2}
\end{align*}
How do I use this to show
$\begin{align*}
\int_{-\infty}^\infty \frac 1{(a^2+s^2)(b^2+s^2)} ds=\frac {\pi}{ab(a+b)}
\end{align*}$
My attempts have been useless, am I supposed to use Parceval's relation or maybe the inversion formula for fourier transforms? I really have no idea.

Using partial fractions we can assume there are numbers $A$ and $B$ where
$$
\frac{1}{(a^2 + s^2)(b^2 + s^2)} \;\; =\;\; \frac{A}{a^2 + s^2} + \frac{B}{b^2 + s^2}.
$$
This is equivalent to writing 
$$
1 \;\; =\;\; A(b^2 + s^2) + B(a^2 + s^2).
$$
If you pick the values of $s = ia$ and $s = ib$ you'll find that $A = -B = \frac{1}{b^2 - a^2}$.  
