What is laplacetransform e^t(t+5)^3 ?

The answer to laplacetransform e^t(t+5)^3 is (125s^3-300s^2+255s-74)/((s-1)^4)

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laplacetransform e^t(t+5)^3

laplace transform

e^{t} (t + 5)^{3}

\frac{125 s ^{3} - 300 s ^{2} + 255 s - 74}{( s - 1 ) ^{4}}

Solution steps

L {e^{t} (t + 5)^{3}}

Apply transform rule:

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

then

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

For

e^{t} (t + 5)^{3} : f (t) = (t + 5)^{3}, a = 1

= L {(t + 5)^{3}} (s - 1)

L {(t + 5)^{3}} : \frac{6}{s ^{4}} + \frac{30}{s ^{3}} + \frac{75}{s ^{2}} + \frac{125}{s}

= \frac{6}{( s - 1 ) ^{4}} + \frac{30}{( s - 1 ) ^{3}} + \frac{75}{( s - 1 ) ^{2}} + \frac{125}{( s - 1 )}

= \frac{6}{( s - 1 ) ^{4}} + \frac{30}{( s - 1 ) ^{3}} + \frac{75}{( s - 1 ) ^{2}} + \frac{125}{s - 1}

Simplify

\frac{6}{( s - 1 ) ^{4}} + \frac{30}{( s - 1 ) ^{3}} + \frac{75}{( s - 1 ) ^{2}} + \frac{125}{s - 1} : \frac{125 s ^{3} - 300 s ^{2} + 255 s - 74}{( s - 1 ) ^{4}}

= \frac{125 s ^{3} - 300 s ^{2} + 255 s - 74}{( s - 1 ) ^{4}}

\frac{\partial ^{2}}{\partial h ( w ) \partial w} (\frac{w}{h ( w ) ( w ) ^{2}})

\frac{5 e ^{- s}}{4 s ^{2} + 36}

- 2 cos (\frac{x}{2})

x \to \infty lim (arctan (4 x))

f (x) = ln (3 x + 4), (2, ln (10))

What is laplacetransform e^t(t+5)^3 ?
The answer to laplacetransform e^t(t+5)^3 is (125s^3-300s^2+255s-74)/((s-1)^4)