What is inverse oflaplace ((s+2)^2)/((s+1)^3) ?

The answer to inverse oflaplace ((s+2)^2)/((s+1)^3) is e^{-t}+e^{-t}*2t+(e^{-t}t^2)/2

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inverse oflaplace ((s+2)^2)/((s+1)^3)

inverse laplace

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

e^{- t} + e^{- t} \cdot 2 t + \frac{e ^{- t} t ^{2}}{2}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

L^{- 1} {\frac{2}{( s + 1 ) ^{2}}} : e^{- t} \cdot 2 t

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

= e^{- t} + e^{- t} \cdot 2 t + \frac{e ^{- t} t ^{2}}{2}

\int 10 x - x e^{x} + e d x

\frac{( 2 x ^{2} + 4 )}{2 x - 2}

\int b x^{3} d x

\frac{x}{1 + x}

\frac{s}{2 s ^{2} + 1}

What is inverse oflaplace ((s+2)^2)/((s+1)^3) ?
The answer to inverse oflaplace ((s+2)^2)/((s+1)^3) is e^{-t}+e^{-t}*2t+(e^{-t}t^2)/2