Inverse Laplace transform $\mathscr{L}^{-1}\{ \frac{1-1e^{-2s}}{s(s+2)} \}$

Question

I am trying to calculate the following inverse Laplace transform. I tried to apply partial fraction decomposition to make it easier to take the inverse but it doesn't seem to work, $s$ is a power in the numerator.

$$\mathscr{L}^{-1}\left\{ \frac{1-1e^{-2s}}{s(s+2)} \right\}$$

Answer 1

$$\mathscr{L}^{-1}\left\{ \frac{1-e^{-2s}}{s(s+2)} \right\}=\mathscr{L}^{-1} \left\{ \frac{1}{s(s+2)} -\frac{e^{-2s}}{s(s+2)}\right\}\\=\mathscr{L}^{-1}\left\{ \frac{1/2}{s}-\frac{1/2}{s+2} -\frac{e^{-2s}}{s(s+2)}\right\}\\=\frac{1}2\mathscr{L}^{-1}\{1/s\}-\frac{1}{2}\mathscr{L}^{-1}\{1/(s+2)\}-\mathscr{L}^{-1}\left\{\frac{e^{-2s}}{s(s+2)}\right\}$$ and $$\mathscr{L}^{-1} \left\{ \frac{1}{s(s+2)}\right\}=\frac{1}2\mathscr{L}^{-1}\{1/s\}-\frac{1}{2}\mathscr{L}^{-1}\{1/(s+2)\}=\frac{1}2(1)-\frac{1}2(e^{-2t})=f(t)$$ and $$\mathscr{L}^{-1}\{e^{-as}F(s)\}=f(t-a)\mathscr{U}(t-a),~~a>0$$ where, here, $a=2$.

Answer 2

You can write $$ \frac{1-1e^{-2s}}{s(s+2)}=\frac{1}{2}\left[A(s)-A(s)e^{-2s}\right] $$ with $$ A(s)=\frac{1}{s}-\frac{1}{s+2} $$ So you can find $$ a(t)=\mathcal{L}^{-1}(A(s))=1+e^{-2t} $$ and $$ \mathcal{L}^{-1}(A(s)e^{-2s})=u(t-2)a(t-2) $$