The inverse Laplace transform of the signal X(s) = log(s + 3) - log(s + 2). Then x(t) is equal to (Assume all initial conditions are zero).
1
\(-\rm \left( \frac{e^{3t} - e^{-2t}}{t} \right) u(t)\)
2
\(-\rm \left( \frac{e^{-3t} - e^{-2t}}{t} \right) u(t)\)
3
\(\rm \left( \frac{e^{-3t} - e^{2t}}{t} \right) u(t)\)
4
\(\rm \left( \frac{e^{3t} - e^{2t}}{t} \right) u(t)\)