Inverse Laplace theorem L^-1{s/((s^2+a^2)(s^2+b^2))

Given,

$L^{-1}\frac{s^2}{(s^2+a^2)(s^2+b^2)} =\ ?$

Let us convert the given function into partial fractions.

$\implies L^{-1}{[\frac{s^2}{(s^2+a^2)(s^2+b^2)}]}$

$\implies L^{-1}[\frac{a^2}{a^2-b^2} \cdot \frac{1}{s^2+a^2}\ -\ \frac{b^2}{a^2-b^2} \cdot \frac{1}{s^2+b^2}]$

$\implies \frac{1}{a^2 - b^2}\ L^{-1}\ [\frac{a^2}{s^2+a^2} -\ \frac{b^2}{s^2-b^2} ]$

$\implies \frac{1}{a^2 - b^2}\ [a^2(\frac{1}{a} sin\ at) -\ b^2(\frac{1}{b} sin\ bt)]$

$\implies \frac{1}{a^2 - b^2}\ [a\ sin\ at - b\ sin\ bt].$

$\therefore L^{-1}\frac{s^2}{(s^2+a^2)(s^2+b^2)} = \frac{1}{a^2 - b^2}\ [a\ sin\ at - b\ sin\ bt].\ Ans.$