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

The answer to inverse oflaplace (s+1)/(s(s+2)(s+4)) is 1/8 \H(t)+1/4 e^{-2t}-3/8 e^{-4t}

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

inverse laplace

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

\frac{1}{8} H (t) + \frac{1}{4} e^{- 2 t} - \frac{3}{8} e^{- 4 t}

Solution steps

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

Take the partial fraction of

\frac{s + 1}{s ( s + 2 ) ( s + 4 )} : \frac{1}{8 s} + \frac{1}{4 ( s + 2 )} - \frac{3}{8 ( s + 4 )}

= L^{- 1} {\frac{1}{8 s} + \frac{1}{4 ( s + 2 )} - \frac{3}{8 ( s + 4 )}}

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}{8 s}} + L^{- 1} {\frac{1}{4 ( s + 2 )}} - L^{- 1} {\frac{3}{8 ( s + 4 )}}

L^{- 1} {\frac{1}{8 s}} : \frac{1}{8} H (t)

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

L^{- 1} {\frac{3}{8 ( s + 4 )}} : \frac{3}{8} e^{- 4 t}

= \frac{1}{8} H (t) + \frac{1}{4} e^{- 2 t} - \frac{3}{8} e^{- 4 t}

\frac{1}{s ^{2} ( s - 3 ) ^{2}}

\int \frac{1}{t - 2} d t

\int \frac{3 x ^{3}}{4 - x ^{2}} d x

\int \frac{1}{x ^{n}} d x

\int t^{3} e^{- 2 t} d t

What is inverse oflaplace (s+1)/(s(s+2)(s+4)) ?
The answer to inverse oflaplace (s+1)/(s(s+2)(s+4)) is 1/8 \H(t)+1/4 e^{-2t}-3/8 e^{-4t}