What is laplacetransform e^{-t}(t-1) ?

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

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laplacetransform e^{-t}(t-1)

laplace transform

e^{- t} (t - 1)

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

Solution steps

L {e^{- t} (t - 1)}

Apply transform rule:

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

then

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

For

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

= L {(t - 1)} (s + 1)

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

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

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

Simplify

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

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

\int \frac{x}{6 x - 5} d x

y^{^{'}} = 3 y^{\frac{2}{3}}

\frac{d}{d y} (e^{- x} + e^{- 2 y})

f (x) = sin^{4} (x)

\frac{d}{d x} (100 x^{2})

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