# 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.

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 $$\mathcal{L}^{-1}\left[\frac{8}{(s^2+4)^2}\right]=\frac12(\sin 2t-2t\cos 2t).$$