<s>[INST] Inverse Laplace Transform -s domain How can I find the inverse Laplace transforms of the following function?
$$ G\left(s\right)=\frac{2(s+1)}{s(s^2+s+2)}  $$
I solved so far. After that, how do I do?
$$ \frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}=G\left( s \right)$$
 [/INST] Use: 
$$
  \mathcal{LT}_s\left( \sin(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \sin(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{\alpha}{(s+b)^2 + \alpha^2}
$$
$$
  \mathcal{LT}_s\left( \cos(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \cos(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{b+s}{(s+b)^2 + \alpha^2}
$$
Completing the square: $s^2+s+2 = \left(s+\frac{1}{2}\right)^2 + \frac{7}{4}$. Therefore, decompose the image of Laplace transform accordingly:
$$
   \frac{2 (s+1)}{s
   \left(s^2+s+2\right)}=\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right)
   ^2+\frac{7}{4}}+\frac{3}{\sqrt{7}}\frac{\sqrt{7}/2}{\left(\left(s+\frac{1}{2}\right)^2+\frac{7}{4}\right)}
$$
Compare with the answer by WolframAlpha.
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