What is inverse oflaplace 1/(s^2+2s+10) ?

The answer to inverse oflaplace 1/(s^2+2s+10) is e^{-t} 1/3 sin(3t)

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inverse oflaplace 1/(s^2+2s+10)

inverse laplace

\frac{1}{s ^{2} + 2 s + 10}

e^{- t} \frac{1}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 2 s + 10}}

\frac{1}{s ^{2} + 2 s + 10} = \frac{1}{( s + 1 ) ^{2} + 9}

= L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 9}}

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} + 9} : a = - 1, F (s) = \frac{1}{s ^{2} + 9}

= e^{- t} L^{- 1} {\frac{1}{s ^{2} + 9}}

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= e^{- t} \frac{1}{3} sin (3 t)

\int \frac{5 x ^{2}}{( x - 2 ) ( x ^{2} + 4 )} d x

\frac{d}{d x} (x + 1 \cdot e^{x})

\frac{7}{x ^{2}}

\int tan^{5} (4 x) d x

\int \frac{x}{3 x ^{2} + 8 x - 3} d x

What is inverse oflaplace 1/(s^2+2s+10) ?
The answer to inverse oflaplace 1/(s^2+2s+10) is e^{-t} 1/3 sin(3t)