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

x \to 5 lim (- 5 x^{2})

i = 1 \sum \infty (\frac{3}{2})^{i}

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

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