What is laplacetransform t^3+3t^2+3t+1 ?

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

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laplacetransform t^3+3t^2+3t+1

laplace transform

t^{3} + 3 t^{2} + 3 t + 1

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

Solution steps

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

e^{\frac{x}{y}} = 3 x - y

\frac{\partial}{\partial x} (1 - x^{2})

\frac{\partial}{\partial y} (x^{4} sin (x y^{3}))

\int \frac{8 + x}{8 - x} d x

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