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?
Using a table, note the form:

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

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

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

Omit $s$
Then we get $\frac d{dT}[L $ inverse of $\frac 16{(s^2+4)^2}]$
Take $16$(const out)
Take $l$ inverse and differentiate with respect to $t$