Solution:

Convolution Theorem:

$Let {\ L[f}_1(t)] = \phi_1(s) \ and \ {\ L[f}_2(t)] = \phi_2(s) \\[2ex] Then, \\ L^{-1}[\phi_1(s).\phi_2 (s)] = \int_0^tf_1(t).f_2(t-u) \ du \\[2ex] where f_1(t)=L^{-1}[\phi_1(s)] \\[2ex] and \ f_2(t)=L^{-1}[\phi_2(s)]$

$L^{-1}\dfrac{{\left(s+3\right)}^2}{{\left(s^2+6s+5\right)}^2}$

$=L^{-1}\dfrac{{\left(s+3\right)}^2}{{\left[{\left(s+3\right)}^2-2^2\right]}^2}$

$e^{-3t}L^{-1}\dfrac{{\left(s\right)}^2}{{\left[{\left(s\right)}^2-2^2\right]}^2}$

By convolution theorem,

$\phi_1(s)=\dfrac{s}{s^2-2^2} \ and \ \phi_2(s)=\dfrac{s}{s^2-2^2}$

$\therefore \ f_1(u)=L^{-1}[\phi_1(s)]=\cosh{2u} \\ and \ f_2(u)=L^{-1}[\phi_2(s)]=\cosh{2u}$

$L^{-1}[\phi(s)] = L^{-1}\left[\dfrac{{\left(s+3\right)}^2}{{\left(s^2+6s+5\right)}^2}\right] = \int_0^t\cosh{2u}.\cosh{2(t-u)}du$

$\dfrac{1}{2}\int_0^t\left[\cosh{2t}+\cosh{2(2u-t)}\right]du$

$\dfrac{1}{2}{\left[u\cosh{2t}+\ \dfrac{sinh\ 2(2u-t)}{4}\right]}_0^t$

$\dfrac{1}{2}\left[t\cosh{2t}+\ \dfrac{1}{4}\sinh{2t}+\dfrac{1}{4}\sinh{2t}\right]$

$\dfrac{\ 1}{4}\left[2tcosh\ 2t+\ \sinh{2t\ }\right]$

$\therefore L^{-1}\dfrac{{\left(s+3\right)}^2}{{\left(s^2+6s+5\right)}^2} = \dfrac{1}{4}e^{-3t}\left[2t\cosh{2t}+\ \sinh{2t}\right]$