Answer
$2 e^{-t} \cos (2t)$
Work Step by Step
Since, $\mathcal{L}(e^{at} \cos bt)=\dfrac{s-a}{(s-a)^2+b^2}$
Here, we have $F(S)=\dfrac{2s+2}{s^2+2s+5}$
or, $=2 \times \dfrac{s+1}{(s+1)^2+(2)^2}$
or, $=2\mathcal{L}(e^{-t} \cos (2t))$
or, $=\mathcal{L}[2e^{-t} \cos (2t)]$
Therefore, the required inverse Laplace transform is:
$\mathcal{L}^{-1}[F(s)]=2 e^{-t} \cos (2t)$
