The correct answer is A. e2tu(t) – 2e-tu(t).
The Laplace transform of a continuous-time signal $x(t)$ is defined as
$$X(s) = \int_{0}^{\infty} x(t) e^{-st} dt$$
The Fourier transform of a continuous-time signal $x(t)$ is defined as
$$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt$$
The Laplace transform and Fourier transform are related by the following equation:
$$X(s) = \int_{0}^{\infty} X(\omega) e^{s\omega} d\omega$$
In this case, we are given that
$$X(s) = {{5 – s} \over {{s^2} – s – 2}}$$
We can use the partial fraction expansion to decompose $X(s)$ as follows:
$$X(s) = {1 \over 2} \left( {1 \over s – 1} + {2 \over s + 2} \right)$$
The inverse Laplace transform of $1 / (s – 1)$ is $e^{t}$, and the inverse Laplace transform of $2 / (s + 2)$ is $2e^{-2t}$. Therefore,
$$x(t) = {1 \over 2} \left( e^{t} + 2e^{-2t} \right)$$
We can write this as
$$x(t) = e^{t} – e^{-2t} + e^{-2t} = e^{2t} – 2e^{-t}$$
Therefore, the correct answer is A.