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

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

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

inverse laplace

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

e^{- t} t

Solution steps

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

\frac{1}{s ^{2} + 2 s + 1} = \frac{1}{( s + 1 ) ^{2}}

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

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

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

= e^{- t} t

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

f (t) = - 2

x \to 0 lim (\frac{0}{x ^{6}})

y = x^{2} + 2 x

x \to 0 lim (x ln (x^{2}))

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