Laplace for a function I have problems to understand Laplace for:
$$\frac{s-1}{(s^2+4)^2}$$
I found that fraction can be wrote with Laplace like this:
$$\frac{s-1}{(s^2+4)^2}=A\cdot\frac{s}{s^2+4}+B\cdot\frac{2}{s^2+4}+C\cdot\frac{4s}{(s^2+4)^2}+D\cdot\frac{s^2-4}{(s^2+4)^2}$$
Can someone please explain me from what are those elements at numerator?

Sal, Alistar .
Just break your initial expression in two pieces  and make use of a Laplace Transform table. The identities that encapsulate our particular case are :

$$  \bullet  \mathcal{L}[t\sin(at)] = \frac{2as}{(s^2+a^2)^2} $$
  $$ \bullet \mathcal{L}[\sin(at) - at\cos(at)] = \frac{2a^3}{(s^2+a^2)^2} $$

So, in order to get the desired original function, we use a little algebraic manipulation :

$$\frac{s}{(s^2+4)^2}-\frac{1}{(s^2+4)^2} = \frac{1}{4} \cdot\frac{4s}{(s^2+4)^2} + \frac{1}{16}\cdot\frac{16}{(s^2+4)^2} $$

This finally yields :
$$\mathcal{L}^{-1}\bigg[\frac{1}{4} \cdot\frac{4s}{(s^2+4)^2} + \frac{1}{16}\cdot\frac{16}{(s^2+4)^2}\bigg] =  \frac{t\sin(2t)}{4}  - \frac{\sin(2t)-2t\cos(2t)}{16}$$

For a more hands-on approach, one might use a Bromwich contour as a support of integration. Sure that would require a little complex analysis, but it'd be a bit more fun than just scrolling through the entries of a table.
