What is inverse oflaplace s/((s+1)^5) ?

The answer to inverse oflaplace s/((s+1)^5) is (e^{-t}t^3)/6-(e^{-t}t^4)/(24)

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inverse oflaplace s/((s+1)^5)

inverse laplace

\frac{s}{( s + 1 ) ^{5}}

\frac{e ^{- t} t ^{3}}{6} - \frac{e ^{- t} t ^{4}}{24}

Solution steps

L^{- 1} {\frac{s}{( s + 1 ) ^{5}}}

Take the partial fraction of

\frac{s}{( s + 1 ) ^{5}} : \frac{1}{( s + 1 ) ^{4}} - \frac{1}{( s + 1 ) ^{5}}

= L^{- 1} {\frac{1}{( s + 1 ) ^{4}} - \frac{1}{( s + 1 ) ^{5}}}

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 ) ^{4}}} - L^{- 1} {\frac{1}{( s + 1 ) ^{5}}}

L^{- 1} {\frac{1}{( s + 1 ) ^{4}}} : \frac{e ^{- t} t ^{3}}{6}

L^{- 1} {\frac{1}{( s + 1 ) ^{5}}} : \frac{e ^{- t} t ^{4}}{24}

= \frac{e ^{- t} t ^{3}}{6} - \frac{e ^{- t} t ^{4}}{24}

2 sin (4 x)

\int sin^{3} (3 x) cos^{5} (3 x) d x

\int_{- 2}^{14} \frac{1}{( x + 2 ) ^{\frac{1}{4}}} d x

\int \frac{x ^{3}}{x} d x

\int x + 2 - x^{2} d x

What is inverse oflaplace s/((s+1)^5) ?
The answer to inverse oflaplace s/((s+1)^5) is (e^{-t}t^3)/6-(e^{-t}t^4)/(24)