# Partial Fractions - Denominator Decomposition dealing with $(ax + b)^k$

So, recently I've been learning about partial fractions and their decomposition and use in Integration, however, I just don't understand why I should be decomposing some fractions in the way that they've told me to. Take the example

$$\frac{1}{(3x + 4)^5}.$$

To decompose this, I would have to convert this to

$$\frac{A}{3x + 4} + \frac{B}{(3x + 4)^2} + \frac{C}{(3x + 4)^3} + \frac{D}{(3x + 4)^4} + \frac{E}{(3x + 4)^5}.$$

Now, the next step would be to equate the numerators, but I don't see how this is possible when the denominators don't line up with each other, as the multiplication of the all the partial sum's denominators would be equal to $(3x + 4)^{15}$ and not $(3x + 4)^5.$

• are you sure all the denominators would be multiplied?Dont forget that we take the LCM of the the denominators which in this case is(3x+4)^5 only Oct 8, 2016 at 1:42

Unless I misunderstand the use of partial fractions, there isn't much you can do to this. The general use of partial fractions is to transform

$$\frac{1}{P(x)Q(x)} = \frac{A}{P(x)}+\frac{B}{Q(x)}$$

where $P$ and $Q$ are polynomials and $A$ and $B$ are constants. The idea is to reduce the degree of the denominator.

However, you can't factor $P(x)Q(x) = (3x+4)^5$ into two polynomials such that a partial fraction decomposition exists (and you don't need to if you're integrating this - just substitute $u=3x+4$).

As AlgorithmsX put it:

"You can rewrite it just as [you] had it, but you would end up with the original rational function after solving for the coefficients."

• "You can rewrite it just as he had it, but you would end up with the original rational function after solving for the coefficients," is what you need to say. Oct 8, 2016 at 2:02
• @AlgorithmsX I added that into my answer. Oct 8, 2016 at 2:03