What is laplacetransform f(t)=(t+1)^3 ?

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

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laplacetransform f(t)=(t+1)^3

laplace transform

f (t) = (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}

(- 1, - 1), (1, - 4)

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\int sin (π) x cos (π) x d x

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