What is inverse oflaplace (s+3)/(s^2+6s+13) ?

The answer to inverse oflaplace (s+3)/(s^2+6s+13) is e^{-3t}cos(2t)

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inverse oflaplace (s+3)/(s^2+6s+13)

inverse laplace

\frac{s + 3}{s ^{2} + 6 s + 13}

e^{- 3 t} cos (2 t)

Solution steps

L^{- 1} {\frac{s + 3}{s ^{2} + 6 s + 13}}

\frac{s + 3}{s ^{2} + 6 s + 13} = \frac{s + 3}{( s + 3 ) ^{2} + 4}

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

Apply inverse transform rule:

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

then

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

For

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

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

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

= e^{- 3 t} cos (2 t)

\frac{d}{d x} (x \times lo g_{10} (x))

\int tan^{3} (x) sec^{10} (x) d x

\int 6 x (x^{2} + 1)^{\frac{1}{2}} d x

\int (3 x - sin (x)) d x

\frac{\partial}{\partial y} (2 x^{3} y - 3)

What is inverse oflaplace (s+3)/(s^2+6s+13) ?
The answer to inverse oflaplace (s+3)/(s^2+6s+13) is e^{-3t}cos(2t)