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

The answer to inverse oflaplace 1/((s-1)^3) is (e^tt^2)/2

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

inverse laplace

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

\frac{e ^{t} t ^{2}}{2}

Solution steps

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

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 ) ^{3}} : a = 1, F (s) = \frac{1}{s ^{3}}

= e^{t} L^{- 1} {\frac{1}{s ^{3}}}

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

= \frac{e ^{t} t ^{2}}{2}

\frac{d}{d x} ((8 x + e^{- x^{2}})^{3})

f (x) = \frac{6}{x ^{3}}

\frac{d y}{d x} = \frac{11 x}{y}

\frac{d}{d x} (arctan (- \frac{x}{2}))

\frac{d ^{2}}{d x ^{2}} (x^{5})

What is inverse oflaplace 1/((s-1)^3) ?
The answer to inverse oflaplace 1/((s-1)^3) is (e^tt^2)/2