What is laplacetransform e^{-t}t ?

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

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laplacetransform e^{-t}t

laplace transform

e^{- t} t

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

Solution steps

L {e^{- t} t}

Use Laplace Transform table:

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

For

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

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

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

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

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

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

\frac{1}{x - 3}

\int \frac{2 e ^{2 r}}{7 + e ^{2 r}} d r

\frac{d}{d x} (5^{6^{7^{x}}})

e^{x} y \cdot y^{^{'}} = e^{- y} + e^{- 2 x - y}

f (x) = sin (2 x), - \frac{π}{2}, \frac{π}{6}

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