What is laplacetransform-5(t-1)e^t ?

The answer to laplacetransform-5(t-1)e^t is -5(1/((s-1)^2)-1/(s-1))

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laplacetransform-5(t-1)e^t

laplace transform

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

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

Solution steps

L {- 5 e^{t} (t - 1)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= - 5 L {(t - 1) e^{t}}

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

= - 5 (\frac{1}{( s - 1 ) ^{2}} - \frac{1}{s - 1})

\int sin^{7} (x) cos^{7} (x) d x

\int \frac{1}{sin ( x ) + cos ( x ) + 1} d x

\int \frac{6}{x ^{2} + 6 x} d x

\int \frac{1}{x ( ln ( x ^{3} ) )} d x

n = 1 \sum \infty \frac{5}{n ^{2} + 64}

What is laplacetransform-5(t-1)e^t ?
The answer to laplacetransform-5(t-1)e^t is -5(1/((s-1)^2)-1/(s-1))