$L^{-1} \left [ \dfrac {s}{(s^2+4)} \right ] \\ =L^{-1} \left [ \dfrac {s}{(s^2+2^2)} \right ] \\ = \cos 2t $

$\text{By Convolution Theorem,} \\L^{-1} \left [ \dfrac {s^2}{(s^2+4)^2} \right ] $

$ \\ = L^{-1} \left [ \dfrac {s}{(s^2+4) } \times\dfrac {s}{(s^2+4) } \right ] \\ = \cos 2t * \cos 2t \\ = \int \limits_0^t \cos 2u \cos (2(t-u))du $

$ \\ = \dfrac{1}{2}\int \limits_0^t 2\cos 2u \cos (2t-2u)du \\ = \dfrac{1}{2}\int \limits_0^t [\cos (2u+2t-2u) + \cos (2u-2t+2u)]du $

$ \\ = \dfrac{1}{2}\int \limits_0^t [\cos (2t)+\cos (4u-2t)]du \\ = \dfrac{1}{2}[u\cos 2t +\dfrac{\sin(4u-2t)}{4}]_0^t $

$\\ = \dfrac{1}{8}[4u\cos 2t +\sin(4u-2t)]_0^t \\ = \dfrac{1}{8}[4t\cos 2t +\sin 2t -0-\sin(-2t)] $

$\\ = \dfrac{1}{8}[4t\cos 2t +2\sin 2t ] \\ \bbox[3pt,yellow]{ = \dfrac{1}{4}[2t\cos 2t +\sin 2t ]} $