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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

(x 4 - x^{2})^{^{'}}

n = 1 \sum \infty \frac{1 ^{n}}{n ^{7}}

x \to 0 lim (\frac{x ^{2}}{∣ x ∣})

y^{^{'}} = t^{2} y^{2}

f (x) = 2 x^{3} - 6 x^{2}

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