<s>[INST] Which rules are used to make function like one in Laplace Transformations table? I have function like this: 
$$\frac{s^2+3s+3}{(2s^2+7s+7)} $$
It needs to be brought to the level of Laplace Transformations from table, like these two:
$$\frac{a}{(s-b)^2 + a^2} $$
$$\frac{s-b}{(s-b)^2 + a^2} $$
I have the solution for this problem, but I can't figure out the logic behind it. Steps included are:
$$\frac{1}{2} - \frac{0.5s+0.5}{2s^2+7s+7} = \frac{1}{2} - \frac{0.5s+0.5}{(s+1.75)^2 + 0.6614^2} = \frac{1}{2} - 0.5*\frac{s+1.75}{(s+1.75)^2 + 0.6614^2} + (0.5*1.75-0.5)*\frac{1}{0.6614}*\frac{0.6614}{(s+1.75)^2+0.6614^2}$$
 [/INST] $$\begin{align}
\frac{s^2+3s+3}{(2s^2+7s+7)}&=\frac12 \frac{2s^2+6s+6}{(2s^2+7s+7)}\\\\
&=\frac12 \frac{(2s^2+7s+7)-(s+1)}{(2s^2+7s+7)}\\\\
&=\frac12 -\frac12\frac{s+1}{2s^2+7s+7}\\\\
&=\frac12-\frac14\frac{s+1}{s^2+(7/2)s+(7/2)}\\\\
&=\frac12-\frac14\frac{s+1}{(s+7/4)^2+(7/2)-(49/16)}\\\\
& =\frac12-\frac14\frac{s+(7/4)-(3/4)}{(s+7/4)^2+7/16}\\\\
&=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+7/16}+\frac{3}{16}\frac{1}{(s+7/4)^2+7/16}\\\\
&=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+7/16}+\frac{3}{16}\frac{1}{(s+7/4)^2+7/16}\\\\
&=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}+\frac{3}{16}\frac{1}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}\\\\
&\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}+\frac{3\sqrt 7}{28}\frac{\frac{\sqrt 7}{4}}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}
\end{align}$$
</s>