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

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

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

inverse laplace

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

e^{- t} - e^{- t} t

Solution steps

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

Take the partial fraction of

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

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

= e^{- t} - e^{- t} t

\frac{d}{d x} (co tan (x) - tan (x))

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

\frac{d}{d t} (\frac{2 2}{3} t^{\frac{3}{2}})

x \to 0 + lim (\frac{x}{x})

f (x) = e^{3 x + 2}

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