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

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

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

inverse laplace

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

e^{- t} t

Solution steps

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

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

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

= e^{- t} t

\int (2 x + 4) d x

d er i v a t i v e ln (7 x)

t an g e n t f (x) = lo g_{5} (2 x^{2} + 7), at x = 3

\frac{d}{d x} ((\frac{2 x + 1}{e + 1})^{e})

\int_{0}^{5} ∣ x - 2 ∣ d x

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