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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 $(3x + 4)^5.$

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  • $\begingroup$ 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 $\endgroup$ – Amritansh Singhal Oct 8 '16 at 1:42
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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."

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  • $\begingroup$ "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. $\endgroup$ – AlgorithmsX Oct 8 '16 at 2:02
  • $\begingroup$ @AlgorithmsX I added that into my answer. $\endgroup$ – Carl Schildkraut Oct 8 '16 at 2:03

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