What is laplacetransform t(t^2-1)^2 ?

The answer to laplacetransform t(t^2-1)^2 is -(-(120)/(s^6)+(12)/(s^4)-1/(s^2))

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laplacetransform t(t^2-1)^2

laplace transform

t (t^{2} - 1)^{2}

- (- \frac{120}{s ^{6}} + \frac{12}{s ^{4}} - \frac{1}{s ^{2}})

Solution steps

L {t (t^{2} - 1)^{2}}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

t (t^{2} - 1)^{2} : f (t) = (t^{2} - 1)^{2}, k = 1

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

L {(t^{2} - 1)^{2}} : \frac{24}{s ^{5}} - \frac{4}{s ^{3}} + \frac{1}{s}

\frac{d}{d s} (\frac{24}{s ^{5}} - \frac{4}{s ^{3}} + \frac{1}{s}) = - \frac{120}{s ^{6}} + \frac{12}{s ^{4}} - \frac{1}{s ^{2}}

= (- 1)^{1} (- \frac{120}{s ^{6}} + \frac{12}{s ^{4}} - \frac{1}{s ^{2}})

= - (- \frac{120}{s ^{6}} + \frac{12}{s ^{4}} - \frac{1}{s ^{2}})

\int 1 - r^{2} \cdot r d r

\frac{\partial}{\partial x} (ln (1 + x))

\int_{0}^{10} 20 x e^{- 0.6 x} d x

\int \frac{x + 2}{x} d x

x \to 0 lim (\frac{3 ^{x} - 2 ^{x}}{x})

What is laplacetransform t(t^2-1)^2 ?
The answer to laplacetransform t(t^2-1)^2 is -(-(120)/(s^6)+(12)/(s^4)-1/(s^2))