Inverse Laplace Transform?

Question

How would you take the inverse laplace transform of $\frac{3p}{(p^2 + 4)^2} $

I attempted to split it as $\frac{3p}{p^2 + 4} $ x $\frac{1}{p^2 + 4} $

Then i know $\frac{3p}{p^2 + 4} $ has laplace tranform $ 3cos(2t) $ and $\frac{1}{p^2 + 4} $ has laplace transform $\frac{1}{2}sin(2t)$

So via convolution theorem the function i want is:

$$\frac{3}{2} \int_{0}^{t} sin(2s)cos(2(t-s)) \, ds$$

Which im not too sure how to integrate?

Or is there a simpler way?

Answer 1

"Or is there a simpler way?"

HINT:

$$\frac{d}{dp}\left( \frac{-3/2}{p^2+4}\right)=\frac{3p}{(p^2+4)^2}$$