Computing $\mathcal{L}^{-1}\left\{\frac{s}{(s + a)^2 + b^2}\right\}$

Question

How can I compute the inverse Laplace transform $\mathcal{L}^{-1}\left\{\frac{s}{(s + a)^2 + b^2}\right\}$?

I recognize it to somewhat be in the form of a trignometric function, but that hasn't gotten me anywhere.

I'm thinking I'm going to need some sort of clever algebraic manipulation. Partial fractions hasn't gotten me anywhere either. I would appreciate some help with this challenging problem.

Answer 1

You should rewrite it as

$$\mathcal{L}^{-1}\left\{\frac{s}{(s + a)^2 + b^2}\right\}=\mathcal{L}^{-1}\left\{\frac{s+a}{(s + a)^2 + b^2}-\frac{a}{b}\cdot\frac{b}{(s+a)^2 + b^2}\right\} $$

So you get a cosine times an exponential minus a sine.