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

The answer to inverse oflaplace 1/((s(s^2+1))) is \H(t)-cos(t)

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

inverse laplace

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

H (t) - cos (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

= H (t) - cos (t)

\frac{d}{d s} (s^{3})

n = 0 \sum \infty \frac{5 ^{n}}{6 ^{n}}

x^{2} y^{^{'}} + 3 x y = 1

x^{6} - 4 e^{x}

\int \frac{4}{x} + 4 x d x

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