What is laplacetransform (1+e^{-t})^2 ?

The answer to laplacetransform (1+e^{-t})^2 is 1/s+2/(s+1)+1/(s+2)

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laplacetransform (1+e^{-t})^2

laplace transform

(1 + e^{- t})^{2}

\frac{1}{s} + \frac{2}{s + 1} + \frac{1}{s + 2}

Solution steps

L {(1 + e^{- t})^{2}}

Expand

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

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

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

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

L {e^{- 2 t}} : \frac{1}{s + 2}

= \frac{1}{s} + 2 \cdot \frac{1}{s + 1} + \frac{1}{s + 2}

Refine

\frac{1}{s} + 2 \frac{1}{s + 1} + \frac{1}{s + 2} : \frac{1}{s} + \frac{2}{s + 1} + \frac{1}{s + 2}

= \frac{1}{s} + \frac{2}{s + 1} + \frac{1}{s + 2}

y = x^{s i n (x)}

t^{2} - 8 t + 16

\frac{d}{d x} ((2 x)^{6 x})

x \to - \infty lim (x^{2} e^{- 3 x})

x - e^{x}

What is laplacetransform (1+e^{-t})^2 ?
The answer to laplacetransform (1+e^{-t})^2 is 1/s+2/(s+1)+1/(s+2)