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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {\frac{1}{s} - \frac{1}{s + 1} - \frac{1}{( 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}} - 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}} : 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}

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

f (x) = x^{2} - 4, - 3, 1

\int x x d x

\frac{d}{d t} (a cos (2 t) + b sin (2 t))

f (x) = (25 - 0.075 x) x, at x = 100

\int \frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

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