What is inverse oflaplace 6/((s+1)^6) ?

The answer to inverse oflaplace 6/((s+1)^6) is (e^{-t}t^5)/(20)

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inverse oflaplace 6/((s+1)^6)

inverse laplace

\frac{6}{( s + 1 ) ^{6}}

\frac{e ^{- t} t ^{5}}{20}

Solution steps

L^{- 1} {\frac{6}{( s + 1 ) ^{6}}}

Apply inverse transform rule:

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

then

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

For

\frac{6}{( s + 1 ) ^{6}} : a = - 1, F (s) = \frac{6}{s ^{6}}

= e^{- t} L^{- 1} {\frac{6}{s ^{6}}}

L^{- 1} {\frac{6}{s ^{6}}} : \frac{t ^{5}}{20}

= \frac{e ^{- t} t ^{5}}{20}

(4 x + 7)^{3}

\int x x^{2} + 16 d x

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

\frac{d}{d θ} (cos (2 θ))

(x^{2} + 1) (\frac{d y}{d x}) + x y - x = 0

What is inverse oflaplace 6/((s+1)^6) ?
The answer to inverse oflaplace 6/((s+1)^6) is (e^{-t}t^5)/(20)