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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{s}{( s + 1 ) ^{2}} : \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 z} (\frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

\int \frac{sin ( \frac{1}{x} )}{x ^{2}} d x

\int x^{4} 4 + x^{5} d x

x \to 0 + lim (x^{2} + 3)

\frac{\partial}{\partial x} (9 x sin (8 x^{2} y))

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