What is inverse oflaplace (s+1)/((s+6)^2(s+2)) ?

The answer to inverse oflaplace (s+1)/((s+6)^2(s+2)) is 1/16 e^{-6t}+(5e^{-6t}t)/4-1/16 e^{-2t}

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inverse oflaplace (s+1)/((s+6)^2(s+2))

inverse laplace

\frac{s + 1}{( s + 6 ) ^{2} ( s + 2 )}

\frac{1}{16} e^{- 6 t} + \frac{5 e ^{- 6 t} t}{4} - \frac{1}{16} e^{- 2 t}

Solution steps

L^{- 1} {\frac{s + 1}{( s + 6 ) ^{2} ( s + 2 )}}

Take the partial fraction of

\frac{s + 1}{( s + 6 ) ^{2} ( s + 2 )} : \frac{1}{16 ( s + 6 )} + \frac{5}{4 ( s + 6 ) ^{2}} - \frac{1}{16 ( s + 2 )}

= L^{- 1} {\frac{1}{16 ( s + 6 )} + \frac{5}{4 ( s + 6 ) ^{2}} - \frac{1}{16 ( s + 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}{16 ( s + 6 )}} + L^{- 1} {\frac{5}{4 ( s + 6 ) ^{2}}} - L^{- 1} {\frac{1}{16 ( s + 2 )}}

L^{- 1} {\frac{1}{16 ( s + 6 )}} : \frac{1}{16} e^{- 6 t}

L^{- 1} {\frac{5}{4 ( s + 6 ) ^{2}}} : \frac{5 e ^{- 6 t} t}{4}

L^{- 1} {\frac{1}{16 ( s + 2 )}} : \frac{1}{16} e^{- 2 t}

= \frac{1}{16} e^{- 6 t} + \frac{5 e ^{- 6 t} t}{4} - \frac{1}{16} e^{- 2 t}

\frac{8}{3 x + 3}

f (x) = \frac{6 cos ( x )}{7 - 9 cos ( x )}

\int \frac{1 - x + 2 x ^{2} - x ^{3}}{x ( x ^{2} + 1 ) ^{2}} d x

cos (2 x^{2})

\int_{0}^{2} \frac{3 ln ( x )}{x} d x

What is inverse oflaplace (s+1)/((s+6)^2(s+2)) ?
The answer to inverse oflaplace (s+1)/((s+6)^2(s+2)) is 1/16 e^{-6t}+(5e^{-6t}t)/4-1/16 e^{-2t}