What is inverse oflaplace 1/((s+1)^5) ?

The answer to inverse oflaplace 1/((s+1)^5) is (e^{-t}t^4)/(24)

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inverse oflaplace 1/((s+1)^5)

inverse laplace

\frac{1}{( s + 1 ) ^{5}}

\frac{e ^{- t} t ^{4}}{24}

Solution steps

L^{- 1} {\frac{1}{( s + 1 ) ^{5}}}

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 ) ^{5}} : a = - 1, F (s) = \frac{1}{s ^{5}}

= e^{- t} L^{- 1} {\frac{1}{s ^{5}}}

L^{- 1} {\frac{1}{s ^{5}}} : \frac{t ^{4}}{24}

= \frac{e ^{- t} t ^{4}}{24}

n = 0 \sum \infty \frac{1}{2 ^{n + 1}}

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

\frac{d ^{2} y}{d x ^{2}} = x^{2}

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\int_{0}^{l n (3)} \frac{4 e ^{x}}{1 + e ^{2 x}} d x

What is inverse oflaplace 1/((s+1)^5) ?
The answer to inverse oflaplace 1/((s+1)^5) is (e^{-t}t^4)/(24)