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}

\int (9 x e^{x^{2}} + \frac{5}{x} - 5^{x} + 3 x^{2}) d x

x \to - 2 lim (n (x))

\frac{2 x ^{3}}{5 y ^{3}}

\int \frac{1}{x x ^{2} - 100} d x

\frac{\partial}{\partial y} (e^{x y z^{7}} y z^{7})

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