What is laplacetransform e^{-t}cos(4t) ?

The answer to laplacetransform e^{-t}cos(4t) is (s+1)/((s+1)^2+16)

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laplacetransform e^{-t}cos(4t)

laplace transform

e^{- t} cos (4 t)

\frac{s + 1}{( s + 1 ) ^{2} + 16}

Solution steps

L {e^{- t} cos (4 t)}

Apply transform rule:

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

then

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

For

e^{- t} cos (4 t) : f (t) = cos (4 t), a = - 1

= L {cos (4 t)} (s + 1)

L {cos (4 t)} : \frac{s}{s ^{2} + 16}

= \frac{( s + 1 )}{( s + 1 ) ^{2} + 16}

= \frac{s + 1}{( s + 1 ) ^{2} + 16}

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

f (x) = - x^{3} - 4 x - 3, at x = - 1

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

(x^{2} + x) e^{x}

x \to 1 + lim (\frac{1}{1 - x})

What is laplacetransform e^{-t}cos(4t) ?
The answer to laplacetransform e^{-t}cos(4t) is (s+1)/((s+1)^2+16)