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

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

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

inverse laplace

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

sin (t)

Solution steps

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

Use Inverse Laplace Transform table:

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

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

= sin (t)

f (x) = 4 ln (x)

f (x) = \frac{6}{4 x + 1}

x \to - 3.1 lim (f (x))

\frac{d}{d x} (\frac{432}{( 1 + 6 x ) ^{3}})

e^{x}, x = 1

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