Inverse laplace transform of $\dfrac{\alpha s}{s+\beta}$

Question

I want to know the inverse laplace transform of $$\dfrac{\alpha s}{s+\beta}$$

where $\alpha, \beta$ are non-zero constants

I already know the result for $$\dfrac{\alpha }{s+\beta}$$

Which is $\alpha \exp(-\beta)$

I know there is a trick for it, but I cannot remember how to deal with the case when you have an extra $s$ on top.

Answer 1

One possibility: $$ \frac{s}{s + \beta} = 1 - \frac{\beta}{s + \beta} $$