Which rules are used to make function like one in Laplace Transformations table?

Question

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:

$$\frac{1}{2} - \frac{0.5s+0.5}{2s^2+7s+7} = \frac{1}{2} - \frac{0.5s+0.5}{(s+1.75)^2 + 0.6614^2} = \frac{1}{2} - 0.5*\frac{s+1.75}{(s+1.75)^2 + 0.6614^2} + (0.5*1.75-0.5)*\frac{1}{0.6614}*\frac{0.6614}{(s+1.75)^2+0.6614^2}$$

Answer 1

$$\begin{align} \frac{s^2+3s+3}{(2s^2+7s+7)}&=\frac12 \frac{2s^2+6s+6}{(2s^2+7s+7)}\\\\ &=\frac12 \frac{(2s^2+7s+7)-(s+1)}{(2s^2+7s+7)}\\\\ &=\frac12 -\frac12\frac{s+1}{2s^2+7s+7}\\\\ &=\frac12-\frac14\frac{s+1}{s^2+(7/2)s+(7/2)}\\\\ &=\frac12-\frac14\frac{s+1}{(s+7/4)^2+(7/2)-(49/16)}\\\\ & =\frac12-\frac14\frac{s+(7/4)-(3/4)}{(s+7/4)^2+7/16}\\\\ &=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+7/16}+\frac{3}{16}\frac{1}{(s+7/4)^2+7/16}\\\\ &=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+7/16}+\frac{3}{16}\frac{1}{(s+7/4)^2+7/16}\\\\ &=\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}+\frac{3}{16}\frac{1}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}\\\\ &\frac12 -\frac14\frac{s+7/4}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2}+\frac{3\sqrt 7}{28}\frac{\frac{\sqrt 7}{4}}{(s+7/4)^2+\left(\frac{\sqrt 7}{4}\right)^2} \end{align}$$

Answer 2

I have outlined the logic of the calculations step by step below. However, your solution seems to have slight problems.

$$ \begin{align} \frac{s^2+3s+3}{(2s^2+7s+7)} &= \frac{s^2+3s+s/2-s/2+3+1/2-1/2}{(2s^2+7s+7)}\\ &= \frac{s^2+3s+s/2-s/2+3+1/2}{(2s^2+7s+7)} - \frac{s/2+1/2}{(2s^2+7s+7)}\\ &=\frac{1}{2}-\frac{1}{2}\frac{s+1}{(2s^2+7s+7)}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1}{(s^2+7s/2+7/2)}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1}{(s^2+2\times 7/4\times s+7/2)}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1}{(s^2+2\times 7/4\times s+(7/4)^2-(7/4)^2+7/2)}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1}{(s+7/4)^2-(7/4)^2+7/2)}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1}{(s+1.75)^2-(1.75)^2+3.5}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1}{(s+1.75)^2+0.4375}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1}{(s+1.75)^2+(0.6614)^2}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1+0.75-0.75}{(s+1.75)^2+(0.6614)^2}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1.75}{(s+1.75)^2+(0.6614)^2}+\frac{1}{4}\frac{0.75}{(s+1.75)^2+(0.6614)^2}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1.75}{(s+1.75)^2+(0.6614)^2}+\frac{1}{4}\frac{0.75\times 0.6614\div 0.6614}{(s+1.75)^2+(0.6614)^2}\\ &=\frac{1}{2}-\frac{1}{4}\frac{s+1.75}{(s+1.75)^2+(0.6614)^2}+\frac{1.13}{4}\frac{0.6614}{(s+1.75)^2+(0.6614)^2}\\ \end{align} $$