What is laplacetransform 1-e^{-t} ?

The answer to laplacetransform 1-e^{-t} is 1/(s(s+1))

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laplacetransform 1-e^{-t}

laplace transform

1 - e^{- t}

\frac{1}{s ( s + 1 )}

Solution steps

L {1 - e^{- t}}

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} - L {e^{- t}}

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

L {e^{- t}} : \frac{1}{s + 1}

= \frac{1}{s} - \frac{1}{s + 1}

Simplify

\frac{1}{s} - \frac{1}{s + 1} : \frac{1}{s ( s + 1 )}

= \frac{1}{s ( s + 1 )}

cos (x), sin (x), [0, \frac{π}{4}]

\int \frac{7}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\int_{0}^{4} \frac{x}{x ^{2} - 9} d x

\int \frac{5 x + 4}{x - 1} d x

\frac{d ^{2}}{d x ^{2}} (5 e^{x} cos (x))

What is laplacetransform 1-e^{-t} ?
The answer to laplacetransform 1-e^{-t} is 1/(s(s+1))