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.
$$\mathscr{L}^{-1}\left\{ \frac{1-1e^{-2s}}{s(s+2)} \right\}$$
Best Answer
$$\mathscr{L}^{-1}\left\{ \frac{1-e^{-2s}}{s(s+2)} \right\}=\mathscr{L}^{-1} \left\{ \frac{1}{s(s+2)} -\frac{e^{-2s}}{s(s+2)}\right\}\\=\mathscr{L}^{-1}\left\{ \frac{1/2}{s}-\frac{1/2}{s+2} -\frac{e^{-2s}}{s(s+2)}\right\}\\=\frac{1}2\mathscr{L}^{-1}\{1/s\}-\frac{1}{2}\mathscr{L}^{-1}\{1/(s+2)\}-\mathscr{L}^{-1}\left\{\frac{e^{-2s}}{s(s+2)}\right\}$$ 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, $a=2$.