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

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

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

inverse laplace

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

- e^{- t} + cos (t)

Solution steps

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

Take the partial fraction of

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

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

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{s}{s ^{2} + 1}}

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

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

= - e^{- t} + cos (t)

f (x) = \frac{x - 8}{3 x - 4}, at x = 2

\frac{d}{d x} (cos (x z (x)))

\int x^{\frac{4}{3}} - 3 x^{\frac{5}{2}} d x

\int \frac{x ^{3}}{x + 1} d x

\int_{0}^{2} 4 - x^{2} d x

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