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

Determine the Laplace transform of the following integrals
$L\left\{{\int }_{0}^{t}{e}^{-2t}{t}^{2}dt\right\}$

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$L\left\{{\int }_{0}^{t}{e}^{-2t}{t}^{2}dt\right\}$
We know that , if $L\left[f\left(t\right)\right]=F\left(s\right)$
then $L\left[{\int }_{0}^{t}f\left(v\right)dv\right]=\frac{F\left(s\right)}{s}$
Here we have
${\int }_{0}^{t}{e}^{-2t}{t}^{2}dt$
so $f\left(t\right)={e}^{-2t}{t}^{2}$
Since , if $L\left[g\left(t\right)\right]=G\left(s\right)$ then
$L\left[{e}^{at}g\left(t\right)\right]=G\left(s-a\right)$
so it we take $g\left(t\right)={t}^{2}$
$⇒G\left(s\right)=L\left[{t}^{2}\right]=\frac{2}{{s}^{3}}$
and hence
$L\left[{e}^{-2t}{t}^{2}\right]={\left(\frac{2}{{s}^{3}}\right)}_{s\to s-\left(-2\right)}$
$⇒L\left[{e}^{-2t}{t}^{2}\right]=\frac{2}{{\left(s+2\right)}^{3}}$
Then
$L\left[{\int }_{0}^{t}{e}^{-2t}{t}^{2}dt\right]=\frac{\frac{2}{{\left(s+2\right)}^{3}}}{s}$
$⇒L\left[{\int }_{0}^{t}{e}^{-2t}{t}^{2}dt\right]=\frac{2}{s{\left(s+2\right)}^{3}}$