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

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

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

inverse laplace

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

cos (t) + sin (t)

Solution steps

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

Expand

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

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

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

= cos (t) + sin (t)

d er i v a t i v e \frac{9}{x}

x \to 0 - lim (3^{\frac{1}{x}})

\int \frac{x ^{2}}{x x ^{2} - x + 1} d x

\int x cos (\frac{1}{2} x) d x

t an g e n t y = sin (x y) + π, (1, π)

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