What is inverse oflaplace 1/(s(s+1)^4) ?

The answer to inverse oflaplace 1/(s(s+1)^4) is \H(t)-e^{-t}-e^{-t}t-(e^{-t}t^2)/2-(e^{-t}t^3)/6

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inverse oflaplace 1/(s(s+1)^4)

inverse laplace

\frac{1}{s ( s + 1 ) ^{4}}

H (t) - e^{- t} - e^{- t} t - \frac{e ^{- t} t ^{2}}{2} - \frac{e ^{- t} t ^{3}}{6}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

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

= H (t) - e^{- t} - e^{- t} t - \frac{e ^{- t} t ^{2}}{2} - \frac{e ^{- t} t ^{3}}{6}

\int 3 tan (5 x) sec^{2} (5 x) d x

h \to 0 lim (5 x^{2} + 8 y^{2})

n = 0 \sum \infty (2 n)! (\frac{x}{3})^{n}

\frac{1}{( 5 s ^{2} + s )}

\frac{d}{d x} (\frac{x ^{2} + 1}{x - 9})

What is inverse oflaplace 1/(s(s+1)^4) ?
The answer to inverse oflaplace 1/(s(s+1)^4) is \H(t)-e^{-t}-e^{-t}t-(e^{-t}t^2)/2-(e^{-t}t^3)/6