To find: The Laplace transform of the function {L}{leftlbrace{t}^{4}-{t}^{2}-{t}+ sin{sqrt{{{2}{t}}}}rightrbrace}

Approach:
With the help of Laplace Transform table and linearity property of Laplace Transformation calculation is carried out as follows:
Calculation:
$L\{t^4-t^2-t+\sin \sqrt{2t}\}$
Consider $f,f_1\text{and}{f}_2\text{be the functions whose Laplace Transform exists for}s>\alpha$ and c be a constant.
Then, for $s>\alpha$
$L\left\{f_1+f_2\right\}=L\left\{f_1\right\}+L\left\{f_2\right\}$
Also,
$L\left\{cf\right\}=cL\left\{f\right\}$
Using above properties:
$L\{t^4-t^2-t+\sin \sqrt{2t}\}=L\left\{t^4\right\}+L\{-t^2\}+L\{-t\}+L\{\sin \sqrt{2t}\}$
$=L\left\{t^4\right\}-L\left\{t^2\right\}-L\left\{t\right\}+L\{\sin \sqrt{2t}\}$
Since the Laplace Transform from the Laplace Transform table:
$L\left\{t^n\right\}=\frac{n!}{s^{n+1}},\mathrm{\forall }s>0,n=1,2,3,\dots$
$L\{\sin at\}=\frac{a}{s^2+a^2},\mathrm{\forall }s>0$
Using above formulae:
$L\left\{t^4\right\}-L\left\{t^2\right\}-L\left\{t\right\}+L\{\sin \sqrt{2t}\}=\frac{4!}{s^{4+1}}-\frac{2!}{s^{2+1}}-\frac{1!}{s^{1+1}}+\frac{\sqrt{2}}{s^2+{\left(\sqrt{2}\right)}^2},\mathrm{\forall }s>0$
$=\frac{24}{s^5}-\frac{2}{s^3}-\frac{1}{s^2}+\sqrt{/}(s^2+2),\mathrm{\forall }s>0$
Gives, the Laplace Transform is
$F\left(s\right)=\frac{24}{s^5}-\frac{2}{s^3}-\frac{1}{s^2}+\sqrt{/}(s^2+2),\mathrm{\forall }s>0$
Conclusion:
Therefore, the required Laplace Transform for the given function is
$F\left(s\right)=\frac{24}{s^5}-\frac{2}{s^3}-\frac{1}{s^2}+\sqrt{/}(s^2+2),\mathrm{\forall }s>0$