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}

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