What is inverse oflaplace 1/((s+1)^4) ?

The answer to inverse oflaplace 1/((s+1)^4) is (e^{-t}t^3)/6

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inverse oflaplace 1/((s+1)^4)

inverse laplace

\frac{1}{( s + 1 ) ^{4}}

\frac{e ^{- t} t ^{3}}{6}

Solution steps

L^{- 1} {\frac{1}{( s + 1 ) ^{4}}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s + 1 ) ^{4}} : a = - 1, F (s) = \frac{1}{s ^{4}}

= e^{- t} L^{- 1} {\frac{1}{s ^{4}}}

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

= \frac{e ^{- t} t ^{3}}{6}

\frac{\partial}{\partial x} (y e^{6 x})

f (x) = (3) e^{- 5 x}

\int_{0}^{2} 4 t^{7} d t

\int (\frac{x ^{2} - 1}{x ^{\frac{3}{2}}}) d x

\int_{0}^{\frac{π}{6}} sin^{2} (x) d x

What is inverse oflaplace 1/((s+1)^4) ?
The answer to inverse oflaplace 1/((s+1)^4) is (e^{-t}t^3)/6