What is laplacetransform (t-1)^3 ?

The answer to laplacetransform (t-1)^3 is 6/(s^4)-6/(s^3)+3/(s^2)-1/s

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laplacetransform (t-1)^3

laplace transform

(t - 1)^{3}

\frac{6}{s ^{4}} - \frac{6}{s ^{3}} + \frac{3}{s ^{2}} - \frac{1}{s}

Solution steps

L {(t - 1)^{3}}

Expand

(t - 1)^{3} : t^{3} - 3 t^{2} + 3 t - 1

= L {t^{3} - 3 t^{2} + 3 t - 1}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t^{3}} - 3 L {t^{2}} + 3 L {t} - L {1}

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

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

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

L {1} : \frac{1}{s}

= \frac{6}{s ^{4}} - 3 \cdot \frac{2}{s ^{3}} + 3 \cdot \frac{1}{s ^{2}} - \frac{1}{s}

Refine

\frac{6}{s ^{4}} - 3 \frac{2}{s ^{3}} + 3 \frac{1}{s ^{2}} - \frac{1}{s} : \frac{6}{s ^{4}} - \frac{6}{s ^{3}} + \frac{3}{s ^{2}} - \frac{1}{s}

= \frac{6}{s ^{4}} - \frac{6}{s ^{3}} + \frac{3}{s ^{2}} - \frac{1}{s}

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

\frac{x ^{2}}{5 + 2 x}

(x^{2} - 1)^{3}

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

\frac{d}{d x} (- ln^{4} (2 sin^{2} (x)^{3} - 1))

What is laplacetransform (t-1)^3 ?
The answer to laplacetransform (t-1)^3 is 6/(s^4)-6/(s^3)+3/(s^2)-1/s