The correct choice is (d) \frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}
To explain I would say: F (t) = L^-1{F(s)}
= L^-1{\frac{5}{s^2-9}}
= L^-1{\frac{A}{s-3} + \frac{B}{s+3}}
Hence, A = (s-3) \frac{5}{(s-3)(s+3)}|_{s=3} = \frac{5}{6}
And, B = (s+3) \frac{5}{(s-3)(s+3)}|_{s=-3} = -\frac{5}{6}
The inverse Laplace transform is \frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}.