Discussion
The inverse Laplace transform of the function \(F(s)=\frac{1}{s(s+1)}\) is given by
Answer: D
The inverse transform is \(f(t)=1-e^{-t}\).
Split into partial fractions and invert term by term:
\[ \begin{aligned} \frac{1}{s(s+1)} &= \frac{1}{s}-\frac{1}{s+1} \\ \mathcal{L}^{-1} &: 1 - e^{-t} \end{aligned} \]
A quick check: the final value theorem gives \(\lim_{s \to 0} sF(s) = 1\), and \(1-e^{-t}\) does tend to 1.
\(\sin t\) is the inverse of \(1/(s^{2}+1)\), \(e^{-t}\sin t\) the inverse of \(1/((s+1)^{2}+1)\), and \(e^{-t}\) the inverse of \(1/(s+1)\) alone, which is only one of the two partial fractions.
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