What is laplacetransform (1+t)^3 ?

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

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laplacetransform (1+t)^3

laplace transform

(1 + t)^{3}

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

Solution steps

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

Expand

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

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

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

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

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

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

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

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

Refine

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

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

\int 2 e^{2 t} d t

\frac{d}{d x} ((x^{2} + 2 x)^{4} (x^{2} + 3 x^{2} + 7)^{3})

y = x^{2} - 5 x + 3, (2, - 3)

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

\int \frac{225 - x ^{2}}{x} d x

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