What is inverse oflaplace (x^2+1)/(x^2-x-6) ?

The answer to inverse oflaplace (x^2+1)/(x^2-x-6) is \delta(t)-e^{-2t}+2e^{3t}

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inverse oflaplace (x^2+1)/(x^2-x-6)

inverse laplace

\frac{x ^{2} + 1}{x ^{2} - x - 6}

δ (t) - e^{- 2 t} + 2 e^{3 t}

Solution steps

L^{- 1} {\frac{x ^{2} + 1}{x ^{2} - x - 6}}

Take the partial fraction of

\frac{x ^{2} + 1}{x ^{2} - x - 6} : 1 - \frac{1}{x + 2} + \frac{2}{x - 3}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1}{x + 2}} : e^{- 2 t}

L^{- 1} {\frac{2}{x - 3}} : 2 e^{3 t}

= δ (t) - e^{- 2 t} + 2 e^{3 t}

\int ln (1 + x) d x

5 cos (9 x), 5 - 5 cos (9 x)

f (x) = 8 sin^{6} (x)

x \to 4 + lim (\frac{1}{∣ 4 - x ∣})

What is inverse oflaplace (x^2+1)/(x^2-x-6) ?
The answer to inverse oflaplace (x^2+1)/(x^2-x-6) is \delta(t)-e^{-2t}+2e^{3t}