What is inverse oflaplace F(s)=((s+2)^2)/(s^3) ?

The answer to inverse oflaplace F(s)=((s+2)^2)/(s^3) is \H(t)+4t+2t^2

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inverse oflaplace F(s)=((s+2)^2)/(s^3)

inverse laplace

F (s) = \frac{( s + 2 ) ^{2}}{s ^{3}}

H (t) + 4 t + 2 t^{2}

Solution steps

L^{- 1} {\frac{( s + 2 ) ^{2}}{s ^{3}}}

Expand

\frac{( s + 2 ) ^{2}}{s ^{3}} : \frac{1}{s} + \frac{4}{s ^{2}} + \frac{4}{s ^{3}}

= L^{- 1} {\frac{1}{s} + \frac{4}{s ^{2}} + \frac{4}{s ^{3}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{s}} + L^{- 1} {\frac{4}{s ^{2}}} + L^{- 1} {\frac{4}{s ^{3}}}

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

L^{- 1} {\frac{4}{s ^{3}}} : 2 t^{2}

= H (t) + 4 t + 2 t^{2}

y = (x^{2} - 8 x) e^{x}

f (x) = \frac{2 x - 4}{x + 1}

\int \frac{1}{x a x + b} d x

f (x) = \frac{20 x}{x ^{2} - 5}

\frac{2 x ^{2} + 5 x}{x}

What is inverse oflaplace F(s)=((s+2)^2)/(s^3) ?
The answer to inverse oflaplace F(s)=((s+2)^2)/(s^3) is \H(t)+4t+2t^2