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

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

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

inverse laplace

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

\frac{1}{2} e^{- t} - \frac{1}{2} cos (t) + \frac{1}{2} sin (t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

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

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

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

= \frac{1}{2} e^{- t} - \frac{1}{2} cos (t) + \frac{1}{2} sin (t)

\frac{d}{d x} (arccoth (x))

\frac{d}{d x} (ln (arctan (5 x^{4})))

\frac{d}{d x} (lo g_{x} (y))

\int 6 x^{2} - 4 x + 8 d x

\frac{d y}{d x} = (3 + y)^{2}

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