Tough inverse Laplace transform

I know what the solution is to this inverse Laplace transform, I just have NO idea how to get there.

$$\mathcal{L}^{-1}\left(\frac{16s}{\left(s^2+4\right)^2}\right)$$

Basically, my question is what modification do I have to do to the equation above?

Answer

Using a table, note the form:

$$f(t) = t\sin(at)$$

$$F(s) = \frac{2as}{(s^2+a^2)^2}$$

Using this:
$$f(t) = \mathcal{L}^{-1}\left(\frac{16s}{\left(s^2+4\right)^2}\right) = \mathcal{L}^{-1}\left(\frac{2*2*s}{\left(s^2+2^2\right)^2}*4\right) =4t\sin(2t)$$