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

The answer to inverse oflaplace 1/((s-1)^2+1) is e^tsin(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)

f (x) = csc (x), at x = \frac{π}{6}

\frac{d}{d x} (y^{3} + x y^{2} + y)

\frac{( x + 1 ) ( x ^{2} + 1 )}{x - 2}

\int_{- 1}^{1} (6 x^{5} + 3) d x

y = \frac{9 x}{e ^{x}}

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