Answer
$$ \dfrac{5}{4} e^{2t} \sin (4t)$$
Work Step by Step
Since, $F(s)=\dfrac{5}{(s-2)^2+ 16}$
The inverse Laplace transform of function can be expressed as:
$F(t)=L^{-1} [\dfrac{5}{(s-2)^2+ 16} ]\\=\dfrac{5}{4} L^{-1} [\dfrac{4}{(s-2)^2+ 4^2}] $
Now, apply the first shifting Theorem.
$f(t)= \dfrac{5}{4} e^{2t} \sin (4t)$
