Find the inverse Laplace transform of the function: F(s)=frac(2s^3+9s^2+11s+3)((s+1)^3s)

Find the inverse Laplace transform of the function:
$$\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}$$

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Step 1: Given:
$$\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}$$
Step 2 To determine:
Inverse Laplace transform of the given function:=?
Step 3 Method used:
We use the partial fraction method to solve the given function.
First of all,
We find the partial fraction of the given function.
Then use some formulas of inverse Laplace transformations:
$$L^{-1}\left\{\frac{1}{s+1}\right\}=e^{-t}$$
$$L^{-1}\left\{\frac{2}{(s+1)^2}\right\}=2te^{-t}$$
$$L^{-1}\left\{\frac{1}{(s+1)^3}\right\}=\frac{e^{-t}t^2}{2}$$
$$L^{-1}\left\{\frac{3}{s}\right\}=3H(t)$$
Step 4 Solution to the question:
Take the partial fraction of:
$$\displaystyle{\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}=-{\frac{{{1}}}{{{s}+{1}}}}+{\frac{{{2}}}{{{\left({s}+{1}\right)}^{{2}}}}}+{\frac{{{1}}}{{{\left({s}+{1}\right)}^{{3}}}}}+{\frac{{{3}}}{{{s}}}}$$
Now:
Taking Laplace inverse of the resultant function:
$$L^{-1}\left\{-\frac{1}{s+1}+\frac{2}{(s+1)^2}+\frac{1}{(s+1)^3}+\frac{3}{s}\right\}$$
Step 5
Now:
Solving this expression:
$$L^{-1}\left\{-\frac{1}{s+1}+\frac{2}{(s+1)^2}+\frac{1}{(s+1)^3}+\frac{3}{s}\right\}$$
$$=-L^{-1}\left\{\frac{1}{s+1}\right\}+2L^{-1}\left\{\frac{1}{(s+1)^2}\right\}+L^{-1-1}\left\{\frac{1}{(s+1)^3}\right\}+3L^{-1}\left\{\frac{1}{s}\right\}$$
Substituting all the values as:
$$L^{-1}\left\{\frac{1}{s+1}\right\}=e^{-t}$$
$$L^{-1}\left\{\frac{2}{(s+1)^2}\right\}=2te^{-t}$$
$$L^{-1}\left\{\frac{1}{(s+1)^3}\right\}=\frac{e^{-t}t^2}{2}$$
$$L^{-1}\left\{\frac{3}{s}\right\}=3H(t)$$
Hence,
We get:
$$\displaystyle{L}^{{-{1}}}{\left({F}{\left({s}\right)}\right)}={\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}$$
$$\displaystyle=-{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}+{\frac{{{e}^{{-{t}}}{t}^{{2}}}}{{{2}}}}+{3}{H}{\left({t}\right)}$$
So,
This is the Laplace inverse of the given function.