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

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

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laplacetransform te^{-t}sin(t)

laplace transform

t e^{- t} sin (t)

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

Solution steps

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

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

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

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

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

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

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