Q:

Finding the inverse Laplace transform of $\frac{s^2-4s-4}{s^4+8s^2+16}$

$$F(s) = \frac{s^2-4s-4}{s^4+8s^2+16}$$
My work is as follows,
$$\frac{s^2-4s-4}{(s^2+4)^2}=\frac{s^2+4}{(s^2+4)^2}-\frac{8}{(s^2+4)^2}-\frac{4s}{(s^2+4)^2}$$
The inverse laplace of the first term is, $\frac{1}{2} \sin(2t)$
The second one has no direct transform, perhaps using the convolution theorem would do.
The third and final term is -$t\sin(2t)$. Is there a faster way to solve this instead of using the convolution theorem for the second term.

A:

Note that
$$
\mathcal{L}^{-1}\left[\frac{a}{s^2+a^2}\right]=\sin at
$$
and
$$
\mathcal{L}^{-1}\left[\frac{s^2-a^2}{(s^2+a^2)^2}\right]=t\cos at,
$$
then
$$
\frac{a}{s^2+a^2}-a\cdot\frac{s^2-a^2}{(s^2+a^2)^2}=\frac{2a^3}{(s^2+a^2)^2}.
$$
Hence
$$
\mathcal{L}^{-1}\left[\frac{2a^3}{(s^2+a^2)^2}\right]=\sin at-at\cos at
$$
and
$$
\mathcal{L}^{-1}\left[\frac{8}{(s^2+4)^2}\right]=\frac12(\sin 2t-2t\cos 2t).
$$

