H(s)=\frac{s-5}{(s-3)(s-1)} The inverse Laplace transform of H(s) is equal to f*g

$H\left(s\right)=\frac{s-5}{\left(s-3\right)\left(s-1\right)}$
The inverse Laplace transform of H(s) is equal to
$f\cdot g$
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icebox2686zsd
Note that we can write
$F\left(s\right)=\frac{s-5}{s-3}=1-\frac{2}{s-3}$
Then, $f\left(t\right)=\delta \left(t\right)-2{e}^{3t}u\left(t\right)$ and
$h\left(t\right)={\int }_{0}^{t}{e}^{t-{t}^{\prime }}\left(\delta \left({t}^{\prime }\right)-2{e}^{3{t}^{\prime }}\right){dt}^{\prime }$
$={e}^{t}-{e}^{t}\left({e}^{2t}-1\right)$
$=2{e}^{t}-{e}^{3t}$
Note that partial fraction expansion makes things easier. We simply write
$H\left(s\right)=\frac{2}{s-1}-\frac{1}{s-3}$