Source : page $146$ https://archive.org/details/academicalgebraf00bowsrich/page/n5/mode/2up
The question I am asking does not deal with partial fraction decomposition ( as far as I can tell).
In Edward Bowser's Academic Algebra , I find the following exercise ( I chose : $\frac {1+2x}{1-3x} \rightarrow 1+5x+\frac{15x^3}{1-3x}$ as a mere example).
First I observe that if I were told to do the exercise in the reverse sense, that is , from the decomposed form ( sum of terms) to the composed one, I would not encounter much difficulty. This is simply applying the rule governing the addition of fractions.
But the other direction is not as easy as the other one.
(1) how does one know the number of terms into which the fraction can be decomposed?
(2) should one use a method akin to the partial fraction decomposition technique, I mean , should one write something like this :
$\frac {1+2x}{1-3x}= \frac {A}{1-3x} + \frac {B}{1-3x}+\frac {C}{1-3x}$
and then try to find the appropriate numerators $A, B$ and $C$?
(3) finally, under what name is this kind of decomposition known nowadays?

For the example you chose, consider the following process:
$$\frac{1+2x}{1-3x} = \frac{1-3x+5x}{1-3x}=1+\frac{5x}{1-3x}$$
Notice that we have just 'added zero' in the numerator to make this happen. For the next term we do the same.
$$\frac{5x}{1-3x}=\frac{5x-15x^2+15x^2}{1-3x}=\frac{5x(1-3x)+15x^2}{1-3x} = 5x+\frac{15x^2}{1-3x}$$
We could continue if we wanted to:
$$\frac{15x^2}{1-3x}=\frac{15x^2-45x^3+45x^3}{1-3x}=\frac{15x^2(1-3x)+45x^3}{1-3x}=15x^2+\frac{45x^3}{1-3x}$$
You probably now see the pattern. For each successive term, simply subtract and add '3x' times the numerator of the rational term (think about why we use 3x here). Factor out the original numerator and then separate into a new polynomial term and a new rational term. Since how many times you choose to repeat this process are up to you, you can generate series of arbitrary length. Hope this helps!