I approached this problem in two ways and arrived at different answers. Both ways seem logical to me. Are they both correct, or is one flawed? This is the original problem:
$$\mathfrak L^{-1} \lbrace{e^{-s}\over s(s+1)} \rbrace$$
I know that $\mathfrak L^{-1} \lbrace e^{-cs}F(s) \rbrace$ gives $u_c(t)f(t-c)$. I decide to use this idea on my first try.
First I complete the square in the denominator to create something that mimicked the Laplace transform of $e^{at}sinh(bt)$. Here is my work for just this part:
$${1\over (s+\frac 12)^2-\frac 14}$$
I multiplied and divided by $\frac 12$ to make this fit the criteria to take the inverse transform, and my result is this:
$$2e^{-\frac t2}sinh(\frac t2)$$
The numerator is the Heaviside function where $c=1$, and so I make my adjustments in the argument of the $sinh$ function and $e$ so everything matches up. The final result is this:
$$f(t)=2e^{-\frac 12(t-1)}sinh(\frac 12(t-1))u_1(t)$$
Alright, now for the second attempt of the problem. I went the route of partial fraction decomposition.
$${e^{-s}\over s(s+1)} = \frac As+\frac B{(s+1)}$$
Solving for $A$ and $B$ gives $A=e^{-s}$ and $B=-e^{-s}$. Now I take the inverse transform of each piece:
$$\mathfrak L^{-1}\lbrace {e^{-s}\over s}+{-e^{-s}\over s+1}\rbrace $$
The final result for this attempt yields:
$$f(t)=u_1(t)-u_1(t)e^{-(t-1)}$$
I want to know if both of these are correct. If not, which one is flawed and why? If they are both correct, how are they related? Even using the relationship that $sinh(t)={e^t-e^{-t}\over 2}$ I still wasn't able to make the first look like the second.
Notice:
So:
$$\mathcal{L}_{s}^{-1}\left[\frac{e^{-s}}{s(s+1)}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{e^{-s}}{s}-\frac{e^{-s}}{1+s}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{e^{-s}}{s}\right]_{(t)}-\mathcal{L}_{s}^{-1}\left[\frac{e^{-s}}{1+s}\right]_{(t)}=$$ $$\theta(t-1)-e^{1-t}\theta(t-1)=\theta(t-1)\left(1-e^{1-t}\right)$$