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

The answer to inverse oflaplace 1/((s-1)^4) is (e^tt^3)/6

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

inverse laplace

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

\frac{e ^{t} t ^{3}}{6}

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s - 1 ) ^{4}} : a = 1, F (s) = \frac{1}{s ^{4}}

= e^{t} L^{- 1} {\frac{1}{s ^{4}}}

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

= \frac{e ^{t} t ^{3}}{6}

y = 10 + cot^{2} (x)

\frac{\partial}{\partial x} (arctan (x - y))

\frac{d}{d x} (\frac{8 e ^{x}}{2 e ^{x} + 1})

\int_{0}^{\frac{1}{2}} 1 - x^{2} d x

\int (1 - t) cos (t^{2} - 2 t) d t

What is inverse oflaplace 1/((s-1)^4) ?
The answer to inverse oflaplace 1/((s-1)^4) is (e^tt^3)/6