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

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

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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)

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

f^{^{''}} (x) = 2 x

\int_{0}^{l n (2)} \frac{4}{e ^{x} + 3} d x

\frac{d y}{d x} x^{2} + y x = 1, y (6) = 9

\int e^{e} d e

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