Finding the inverse laplace transform of $\frac{3s+4}{s^2-16}$

Question

How would you find the inverse Laplace transformation of $\displaystyle \frac{3s+4}{s^2-16}$ when $s>4$? Thanks!!

I dont really understand what we need to do for this question. Please help

Answer 1

Hint:

Write the partial fraction fraction expansion as:

$$\displaystyle \tag 1 \frac{3s+4}{s^2-16} = \frac{1}{s+4} + \frac{2}{s-4}$$

Now, take the inverse Laplace of each of the terms on the right-hand-side (RHS) of $(1)$.

We have for $s \gt a$:

$$\mathcal{L}^{-1}\left(\frac{1}{s-a}\right) = e^{at}$$

Clear?