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

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

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

inverse laplace

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

e^{- t} cos (t)

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{s + 1}{( s + 1 ) ^{2} + 1} : a = - 1, F (s) = \frac{s}{s ^{2} + 1}

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

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

= e^{- t} cos (t)

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