Answer
$=\frac{1}{4}[1-e^{-2t}(\cos 4t-\frac{3}{2}\sin 4t)]$
Work Step by Step
Given: $F(s)=\frac{2s+5}{s(s^2+4s+20)}$
Using the Convolution Theorem
$L^{-1}[F(s)]=L^{-1}[\frac{2s+5}{s(s^2+4s+20)}]\\
=L^{-1}[\frac{1}{4s}].L^{-1}[\frac{4-s}{4(s^2+4s+20)}]\\
=\frac{1}{4}[L^{-1}(\frac{1}{s}) -L^{-1}(\frac{s+2-6}{(s+2)^2+16})]\\
=\frac{1}{4}[1-e^{-2t}L^{-1}(\frac{s-6}{s^2+16})]\\
=\frac{1}{4}[1-e^{-2t}(\cos 4t-\frac{3}{2}\sin 4t)]$
