Answer
$\dfrac{3}{5} -\dfrac{3}{5} e^{t} \cos (2t) +\dfrac{13}{10} e^t \sin (2t)$
Work Step by Step
Since, $F(s)=\dfrac{2s+3}{s(s^2-2s+5)}=\dfrac{2}{5} [\dfrac{s+3}{[(s-1)^2}+2^2]$
The inverse Laplace transform of function can be expressed as:
$F(t)=L^{-1} [\dfrac{3}{5s}-\dfrac{3(s-1)}{5(s-1)^2+2^2}+\dfrac{13}{5(s-1)^2+2^2}]$
Now, apply the first shifting Theorem.
$f(t)=\dfrac{3}{5} -\dfrac{3}{5} e^{t} \cos (2t) +\dfrac{13}{10} e^t \sin (2t)$
