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

The answer to inverse oflaplace 1/(x^2+x+1) is e^{-t/2} 2/(sqrt(3))sin((sqrt(3)t)/2)

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

inverse laplace

\frac{1}{x ^{2} + x + 1}

e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

Solution steps

L^{- 1} {\frac{1}{x ^{2} + x + 1}}

\frac{1}{x ^{2} + x + 1} = \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}}

= L^{- 1} {\frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} : a = - \frac{1}{2}, F (x) = \frac{1}{x ^{2} + \frac{3}{4}}

= e^{- \frac{1}{2} t} L^{- 1} {\frac{1}{x ^{2} + \frac{3}{4}}}

L^{- 1} {\frac{1}{x ^{2} + \frac{3}{4}}} : \frac{2}{3} sin (\frac{3 t}{2})

= e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

\int (5 + x^{9})^{6} x^{8} d x

\frac{d}{d x} (2 x^{3} cos (3 x))

x \to 1 lim (6 x + a)

\frac{\partial}{\partial x} ((\frac{x}{y}) - yz)

(\frac{x}{x ^{2} + 9})^{^{'}}

What is inverse oflaplace 1/(x^2+x+1) ?
The answer to inverse oflaplace 1/(x^2+x+1) is e^{-t/2} 2/(sqrt(3))sin((sqrt(3)t)/2)