Determine the Laplace transform of f{{left({t}right)}}={t}^{4}-{t}^{2}{e}^{{{3}{t}}}+{2}

Step 1
The given function is $f\left(t\right)=t^4-t^2{e}^{3t}+2$
Find the Laplace transform of $f\left(t\right)=t^4-t^2{e}^{3t}+2$ as shown below.
Step 2
It is known that,
$L\left\{1\right\}=\frac{1}{s}$
$L\left\{t^n\right\}=\frac{n!}{s^{n+1}}$
($L\left\{t\ne \left(at\right)\right\}=\frac{n!}{{(s-a)}^{n+1}}$
Then,
$L\left\{f\left(t\right)\right\}=L\{t^4-t^2{e}^{3t}+2\}$
$=L\left\{t^4\right\}-L\left\{t^2{e}^{3t}\right\}+L\left\{2\right\}$$=L\left\{t^4\right\}-L\left\{t^2{e}^{3t}\right\}+2L\left\{1\right\}$
$=\frac{24}{s^5}-\frac{2}{{(s-3)}^3}+\frac{2}{s}$

Step 3
Therefore, the Laplace transform of the function $f\left(t\right)=t^4-t^2{e}^{3t}+2\text{ is }L\left\{f\left(t\right)\right\}=\frac{24}{s^5}-\frac{2}{{(s-3)}^3}+\frac{2}{s}$