# How to calculate inverse Laplace transform of (7s^2+3s+5)/((s^2−4s+29)(s^2+25))?

How to calculate inverse Laplace transform of $\frac{7{s}^{2}+3s+5}{\left({s}^{2}-4s+29\right)\left({s}^{2}+25\right)}$?
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Marley Stone
Note that
$\frac{As+B}{{s}^{2}+{5}^{2}}=\frac{A/2-iB/10}{s-5i}+\frac{A/2+iB/10}{s+5i}$
has inverse Laplace transform
$\left(A/2-iB/10\right){e}^{5it}+\left(A/2+iB/10\right){e}^{-5it}=A\mathrm{cos}5t+\frac{B}{5}\mathrm{sin}5t.$
Since ${s}^{2}-4s+29=\left(s-2{\right)}^{2}+{5}^{2}$, the other partial fraction admits a similar analysis.