What is inverse oflaplace (s+1)/(s(s^2+4s+4)) ?

The answer to inverse oflaplace (s+1)/(s(s^2+4s+4)) is 1/4 \H(t)-1/4 e^{-2t}+(e^{-2t}t)/2

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inverse oflaplace (s+1)/(s(s^2+4s+4))

inverse laplace

\frac{s + 1}{s ( s ^{2} + 4 s + 4 )}

\frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} + \frac{e ^{- 2 t} t}{2}

Solution steps

L^{- 1} {\frac{s + 1}{s ( s ^{2} + 4 s + 4 )}}

Take the partial fraction of

\frac{s + 1}{s ( s ^{2} + 4 s + 4 )} : \frac{1}{4 s} - \frac{1}{4 ( s + 2 )} + \frac{1}{2 ( s + 2 ) ^{2}}

= L^{- 1} {\frac{1}{4 s} - \frac{1}{4 ( s + 2 )} + \frac{1}{2 ( s + 2 ) ^{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}{4 s}} - L^{- 1} {\frac{1}{4 ( s + 2 )}} + L^{- 1} {\frac{1}{2 ( s + 2 ) ^{2}}}

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

L^{- 1} {\frac{1}{4 ( s + 2 )}} : \frac{1}{4} e^{- 2 t}

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

= \frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} + \frac{e ^{- 2 t} t}{2}

\frac{d}{d x} (x csc (x))

x^{3} + 2 x

\int 9 sin (x) cos (x) d x

(2 e^{y} - x) y^{^{'}} = 1

\int \frac{x ^{2}}{( 225 + x ^{2} ) ^{2}} d x

What is inverse oflaplace (s+1)/(s(s^2+4s+4)) ?
The answer to inverse oflaplace (s+1)/(s(s^2+4s+4)) is 1/4 \H(t)-1/4 e^{-2t}+(e^{-2t}t)/2