Let $Y(s) = \frac{2e^{-s}}{s(s^2 + 3s + 2)}$. Then the inverse Laplace transform is \begin{align} y(t) &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{2e^{s(t - 1)}}{s(s^2 + 3s + 2)}ds\\ &= \lim_{s\to 0}\frac{2e^{s(t - 1)}}{s^2 + 3s + 2} + \lim_{s\to - 1}\frac{2e^{s(t - 1)}}{s(s + 2)} + \lim_{s\to -2}\frac{2e^{s(t - 1)}}{s(s + 1)}\\ &= 1 - 2e^{-(t-1)} + e^{-2(t - 1)}\\ &= (e^{-(t-1)} - 1)^2\\ &\neq e^{-2(t-1)}(e^{t-1} - 1)^2\mathcal{U}(t - 1)\tag{solution} \end{align} The actual solution was obtained from Mathematica. Where did I go astray?
I have, also, come to the conclusion that whenever there is a shift in the exponential $s(t-a)$ where $a\in\mathbb{R}$, we have to multiple by $\mathcal{U}(t-a)$ but I think I should be able to acquire the unit step from the integration, correct?
I found the problem. \begin{align} y(t) &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{2e^{s(t - 1)}}{s(s^2 + 3s + 2)}ds\\ &= \lim_{s\to 0}\frac{2e^{s(t - 1)}}{s^2 + 3s + 2} + \lim_{s\to - 1}\frac{2e^{s(t - 1)}}{s(s + 2)} + \lim_{s\to -2}\frac{2e^{s(t - 1)}}{s(s + 1)} \end{align} Due to the exponential term, we need the real part of $s$ to be $\text{Re}(s)>1$. So this introduces the $\mathcal{U}(t-1)$ \begin{align} &=1\mathcal{U}(t-1) - 2e^{-(t-1)}\mathcal{U}(t-1) + e^{-2(t-1)}\mathcal{U}(t-1)\\ &=\mathcal{U}(t-1)(1-2e^{-(t-1)}+e^{-2(t-1)})\\ &=\mathcal{U}(t-1)(-e^{-(t-1)}+1)^2 \end{align}