What is inverse oflaplace 1/(e(s-1)^3) ?

The answer to inverse oflaplace 1/(e(s-1)^3) is (e^{t-1}t^2)/2

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inverse oflaplace 1/(e(s-1)^3)

inverse laplace

\frac{1}{e ( s - 1 ) ^{3}}

\frac{e ^{t - 1} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{1}{e ( s - 1 ) ^{3}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{e ( s - 1 ) ^{3}} : a = 1, F (s) = \frac{1}{e s ^{3}}

= e^{t} L^{- 1} {\frac{1}{e s ^{3}}}

L^{- 1} {\frac{1}{e s ^{3}}} : \frac{t ^{2}}{2 e}

= \frac{e ^{t - 1} t ^{2}}{2}

sin (π x)

y = 3 + 5 x^{2} - 2 x^{3}

x \to 0 lim ((5 x + 2)^{3})

ln (25 sin^{2} (x))

x \to \infty lim (\frac{( 3 x ^{3} + 12 x ^{2} - 8 x + 1 )}{( x ^{2} + 2 ) ( 3 x - 1 )})

What is inverse oflaplace 1/(e(s-1)^3) ?
The answer to inverse oflaplace 1/(e(s-1)^3) is (e^{t-1}t^2)/2