What is inverse oflaplace arctan(s)+1 ?

The answer to inverse oflaplace arctan(s)+1 is -(sin(t))/t

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inverse oflaplace arctan(s)+1

inverse laplace

arctan (s) + 1

- \frac{sin ( t )}{t}

Solution steps

L^{- 1} {arctan (s) + 1}

Use Inverse Laplace Transform table:

L^{- 1} {F (s)} = - \frac{1}{t} L^{- 1} {F^{'} (s)}

L^{- 1} {arctan (s) + 1} = - \frac{1}{t} L^{- 1} {(arctan (s) + 1)^{'}}

= - \frac{1}{t} L^{- 1} {(arctan (s) + 1)^{^{'}}}

(arctan (s) + 1)^{'} = \frac{1}{s ^{2} + 1}

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

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

= - \frac{1}{t} sin (t)

= - \frac{sin ( t )}{t}

What is inverse oflaplace arctan(s)+1 ?
The answer to inverse oflaplace arctan(s)+1 is -(sin(t))/t