In Problems\(21 - 30\), determine \({\mathcal{L}^{ - 1}}\{ F\} \).

\(F\left( s \right) = \frac{{5{s^2} + 34s + 53}}{{{{\left( {s + 3} \right)}^2}\left( {s + 1} \right)}}\)

Using partial fractions we get:

\(\frac{{5{s^2} + 34s + 53}}{{{{(s + 3)}^2}(s + 1)}} = \frac{A}{{s + 3}} + \frac{B}{{{{(s + 3)}^2}}} + \frac{C}{{s + 1}}\)

This implies that:

\(5{s^2} + 34s + 53 = A\left( {s + 1} \right)\left( {s + 3} \right) + B\left( {s + 1} \right) + C{\left( {s + 3} \right)^2}\)

Using \(s = - 1, - 3,0\), respectively we get:

Put \(s = - 1\) as:

\(\begin{array}{c}5 - 34 + 53 = 4C\\24 = 4C\\C = 6\end{array}\)

Put \(s = - 3\) as:

\(\begin{array}{c}45 - 102 + 53 = - 2B\\ - 4B = - 2\\B = 2\end{array}\)

Put \(s = 0\) as:

\(\begin{array}{c}53 = 3A + B + 9C\\A = \frac{{53 - B - 9C}}{3}\\A = - 1\end{array}\)

Therefore,

\(\frac{{5{s^2} + 34s + 53}}{{{{(s + 3)}^2}(s + 1)}} = - \frac{1}{{s + 3}} + \frac{2}{{{{(s + 3)}^2}}} + \frac{6}{{s + 1}}\)

Find the inverse Laplace transform as:

\(\begin{array}{c}{\mathcal{L}^{ - 1}}\left\{ {\frac{{5{s^2} + 34s + 53}}{{{{(s + 3)}^2}(s + 1)}}} \right\} = {\mathcal{L}^{ - 1}}\left\{ { - \frac{1}{{s + 3}}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{2}{{{{(s + 3)}^2}}}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{6}{{s + 1}}} \right\}\\ = - {\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{s + 3}}} \right\} + 2{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{{{(s + 3)}^2}}}} \right\} + 6{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{s + 1}}} \right\}\\ = - {e^{ - 3t}} + 2t{e^{ - 3t}} + 6{e^{ - t}}\end{array}\)

Therefore, \({\mathcal{L}^{ - 1}}\left\{ F \right\} = - {e^{ - 3t}} + 2t{e^{ - 3t}} + 6{e^{ - t}}\)