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

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

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

laplace transform

e^{- t} (\frac{e ^{t} + e ^{- t}}{2})

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

Solution steps

L {e^{- t} (\frac{e ^{t} + e ^{- t}}{2})}

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

e^{- t} \frac{e ^{t} + e ^{- t}}{2} : f (t) = \frac{e ^{t} + e ^{- t}}{2}, a = - 1

= L {\frac{e ^{t} + e ^{- t}}{2}} (s + 1)

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

= \frac{1}{2 ( ( s + 1 ) - 1 )} + \frac{1}{2 ( ( s + 1 ) + 1 )}

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

f (x) = \frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7

n \to \infty lim (4 n)

\frac{\partial}{\partial x} (2 x^{\frac{1}{2}} y^{\frac{1}{3}})

y^{^{'}} = \frac{y}{2}

\frac{d}{d x} (- \frac{1}{ln ( 10 ) x ^{2}})

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