inverse laplace transform of $s/(s^2+6s+13)$

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Hint: $\dfrac{s}{s^2+6s+13} = \dfrac{s}{(s+3)^2+2^2} = \dfrac{(s+3)-3}{(s+3)^2+2^2} = \dfrac{s+3}{(s+3)^2+2^2} - \dfrac{3}{(s+3)^2+2^2}$.

Now, look at your Laplace Transform table for expressions like $\dfrac{s+a}{(s+a)^2+b^2}$ and $\dfrac{b^2}{(s+a)^2+b^2}$.

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sanbapuo
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Updated on August 01, 2022

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  • sanbapuo
    sanbapuo over 1 year

    Hi can anyone help with this inverse Laplace transform $$s/(s^2+6s+13) $$ I tried to do partial fraction $s+3/(s+3)^2+4 - 2/(s+3)^2+4$, but then I don't know what to do next...

    • anon
      anon about 9 years
      Your partial fraction expansion is incorrect. Where are you getting $s+3$ from? Clearly the denominator has no rational roots, and partial fractions with radical expressions could turn out complicated - hence completing the square is the best thing to try.