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

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

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laplacetransform te^{-t}cos(t)

laplace transform

t e^{- t} cos (t)

- \frac{- s ^{2} - 2 s}{( s ^{2} + 2 s + 2 ) ^{2}}

Solution steps

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

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

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

= (- 1)^{1} \frac{d}{d s} (L {e^{- t} cos (t)})

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

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

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

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

\int 0. 8^{x} d x

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{9} d x

\int_{0}^{5} x + 1 d x

\frac{d y}{d t} = - \frac{t}{y}

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