What is inverse oflaplace 1/((x+4)^2) ?

The answer to inverse oflaplace 1/((x+4)^2) is e^{-4t}t

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inverse oflaplace 1/((x+4)^2)

inverse laplace

\frac{1}{( x + 4 ) ^{2}}

e^{- 4 t} t

Solution steps

L^{- 1} {\frac{1}{( x + 4 ) ^{2}}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( x + 4 ) ^{2}} : a = - 4, F (x) = \frac{1}{x ^{2}}

= e^{- 4 t} L^{- 1} {\frac{1}{x ^{2}}}

L^{- 1} {\frac{1}{x ^{2}}} : t

= e^{- 4 t} t

4 e^{x} cos (x)

ln (2 x) \cdot 4

\frac{d}{d x} (5^{- 4 x})

x \to 0 lim (e^{5 x})

\int_{0}^{\frac{π}{3}} 3 sec^{2} (x) d x

What is inverse oflaplace 1/((x+4)^2) ?
The answer to inverse oflaplace 1/((x+4)^2) is e^{-4t}t