What is laplacetransform f(t)=(1+e^{4t})^2 ?

The answer to laplacetransform f(t)=(1+e^{4t})^2 is 1/s+2/(s-4)+1/(s-8)

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laplacetransform f(t)=(1+e^{4t})^2

laplace transform

f (t) = (1 + e^{4 t})^{2}

\frac{1}{s} + \frac{2}{s - 4} + \frac{1}{s - 8}

Solution steps

L {(1 + e^{4 t})^{2}}

Expand

(1 + e^{4 t})^{2} : 1 + 2 e^{4 t} + e^{8 t}

= L {1 + 2 e^{4 t} + e^{8 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^{4 t}} + L {e^{8 t}}

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

L {e^{4 t}} : \frac{1}{s - 4}

L {e^{8 t}} : \frac{1}{s - 8}

= \frac{1}{s} + 2 \cdot \frac{1}{s - 4} + \frac{1}{s - 8}

Refine

\frac{1}{s} + 2 \frac{1}{s - 4} + \frac{1}{s - 8} : \frac{1}{s} + \frac{2}{s - 4} + \frac{1}{s - 8}

= \frac{1}{s} + \frac{2}{s - 4} + \frac{1}{s - 8}

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

e^{- 2 t - 5}

\frac{\partial}{\partial x} (y \cdot ln (x))

\frac{d}{d x} (s e^{- s x y (s, x)})

\int \frac{x + 1}{x ^{3} - x ^{2}} d x

What is laplacetransform f(t)=(1+e^{4t})^2 ?
The answer to laplacetransform f(t)=(1+e^{4t})^2 is 1/s+2/(s-4)+1/(s-8)