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)

x \to - 2 lim (4 x^{2})

\frac{d}{d x} (x^{2} - 16)

\int (0.7 y^{3} + 10 + 2 y^{- 3}) d y

f (x) = x^{2} - 2 x - 3

\frac{d}{d x} (sin (cot (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)