What is inverse oflaplace 1/((s^2)) ?

The answer to inverse oflaplace 1/((s^2)) is t

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inverse oflaplace 1/((s^2))

inverse laplace

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

t

Solution steps

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

Use Inverse Laplace Transform table:

L^{- 1} {\frac{( n ) !}{s ^{n + 1}}} = t^{n}

L^{- 1} {\frac{1}{s ^{2}}} = t

= t

\int \frac{1}{x ^{2} - 4 x + 20} d x

f (x) = x^{2} sin (\frac{1}{x})

\frac{\partial}{\partial x} (e^{x + y} sin (z) + (y + 9) arccos (x))

(1 + e^{4 t})^{2}

\int x^{2} 1 + 4 x^{2} d x

What is inverse oflaplace 1/((s^2)) ?
The answer to inverse oflaplace 1/((s^2)) is t