Determine the Laplace transform of the following integrals. L\{\int_0^t e^{-2t}t^2 dt\}

$L\left\{{\int }_0^t{e}^{-2t}{t}^{2}dt\right\}$
We know that , if ${\displaystyle L\left[f\left(t\right)\right]=F\left(s\right)}$
then ${\displaystyle L\left[{\int }_0^{t}f\left(v\right)dv\right]=\frac{F\left(s\right)}{s}}$
Here we have
${\displaystyle {\int }_0^t{e}^{-2t}{t}^{2}dt}$
so ${\displaystyle f\left(t\right)=e^{-2t}{t}^2}$
Since , if ${\displaystyle L\left[g\left(t\right)\right]=G\left(s\right)}$ then
${\displaystyle L\left[e^{at}g\left(t\right)\right]=G(s-a)}$
so it we take ${\displaystyle g\left(t\right)=t^2}$
${\displaystyle \Rightarrow G\left(s\right)=L\left[t^2\right]=\frac{2}{s^3}}$
and hence
${\displaystyle L\left[e^{-2t}{t}^2\right]={\left(\frac{2}{s^3}\right)}_{s\to s-(-2)}}$
${\displaystyle \Rightarrow L\left[e^{-2t}{t}^2\right]=\frac{2}{{(s+2)}^3}}$
Then
${\displaystyle L\left[{\int }_0^t{e}^{-2t}{t}^{2}dt\right]=\frac{\frac{2}{{(s+2)}^3}}{s}}$
${\displaystyle \Rightarrow L\left[{\int }_0^t{e}^{-2t}{t}^{2}dt\right]=\frac{2}{s{(s+2)}^3}}$