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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

= e^{- t} cos (t) - 1 \cdot e^{- t} sin (t)

Refine

e^{- t} cos (t) - 1 e^{- t} sin (t) : e^{- t} cos (t) - e^{- t} sin (t)

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

\frac{d}{d x} ((x - x + 6 x^{4})^{3})

\int (x^{3} (1 + x^{2}))^{- 1} d x

(t + 6)^{3}

h \to 0 lim (- \frac{5 h}{x ( x + h )})

\frac{x ^{3}}{7}

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