This is my textbook from what I am studying and the “green” highlighted part is where I am questioning myself.
Let’s pull out the expression-
$$\mathcal{L} ( \frac{ \frac{19}{25} (s) + \frac{14}{25} }{ (s+2)^2 + 3^2} )$$
From here, I distribute it to become
$ \frac{19}{25} \mathcal{L} ( \frac{s}{ (s+2)^2 + 3^2} )+ \frac{14}{25} \mathcal{L} (\frac{1}{ (s+2)^2 + 3^2 }) $
Where does the $ \mathcal {L} \frac{2}{3} \frac{3}{ (s+2)^2 + 3^2 } $ (green highlighted portion) come from ?
Let us go back one line, then we have three parts:
$$\left\{\frac{6}{25}\frac{1}{s-2}\right\} + \left\{\frac{\frac{19s}{25}}{(s+2)^2+3^2}\right\} + \left\{\frac{\frac{14}{25}}{(s+2)^2+3^2}\right\} $$
Now let us focus on the second part (containing orange and green parts):
$$\left\{\frac{\frac{19s}{25}}{(s+2)^2+3^2}\right\} = \frac{19}{25}\frac{s}{(s+2)^2+3^2} $$
And adding a zero to the nomerator will not change anything:
$$\frac{19}{25}\frac{s+(2-2)}{(s+2)^2+3^2} = \frac{19}{25}\frac{s+2}{(s+2)^2+3^2}+\frac{19}{25}\frac{-2}{(s+2)^2+3^2}$$
Or:
$$\frac{19}{25}\left\{\frac{s+2}{(s+2)^2+3^2}+\frac{2}{3}\frac{-3}{(s+2)^2+3^2}\right\}$$