Inverse Laplace Transform of exponential fraction

Question

I need to inverse Laplace transform this function: $$\frac{1-e^{-2s}}{s(s+a)}$$ (where a is any real, positive constant)

So far I only have that it is equivalent to:

$$\frac{1}{s(s+a)} - \frac{e^{-2s}}{s(s+a)}$$ I'm able to determine the inverse transform of the first term just using partial fractions and standard transform result tables, but I'm struggling with the exponential fraction and am not sure where to begin.

Any help would be much appreciated. Thanks.

Answer 1

The solution is:

$$\frac{1}{a} - \frac{e^{-at}}{a} + \left(\frac{1}{a} - \frac{e^{-a(t-2)}}{a}\right) H(t-2)$$

Where $H(t)$ is the Heaviside step function.

Wolfram Alpha Solution