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

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

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inverse oflaplace 3/((s+1)^3)

inverse laplace

\frac{3}{( s + 1 ) ^{3}}

\frac{3 e ^{- t} t ^{2}}{2}

Solution steps

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

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

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

= \frac{3 e ^{- t} t ^{2}}{2}

f (x) = \frac{x}{1 + x}

f (x) = 4 x^{3}

\frac{d x}{d t} = x + e^{t}

\frac{d}{d x} (3 x + \frac{2}{5 x ^{3}} - 2)

\int sin^{4} (t) d t

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