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

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

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

inverse laplace

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

e^{- t} sin (t)

Solution steps

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

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

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

= e^{- t} sin (t)

\frac{d}{d x} (\frac{x ^{4} + 2}{ln ( x )})

\int \frac{5}{x 16 x ^{2} - 81} d x

8 ln (\frac{5 x + 7}{3 x ^{2} - 6 x})

f (x) = sin (x^{2} + 1)

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