# Partial Fraction Decomposition of $\frac{x^4+2}{x^5+6x^3}$

I tried answering the following question but I'm getting it wrong for some reason. I would appreciate any help.

$$\frac{x^4+2}{x^5+6x^3}$$

$$\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{Dx+E}{x^3}+\frac{Fx+G}{x^2+6}$$

What am I doing wrong?

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How did you get $x+5$ and $x-5$? Can you show your working? –  Aryabhata Mar 4 '12 at 2:57
Sorry I copied the wrong answer from my notepad. Just fixed it –  mathisnotmyforte Mar 4 '12 at 3:00
link and press show steps near "Partial fraction expansion" –  Salech Alhasov Mar 4 '12 at 3:02

The typical way to deal with $(x+b)^n$ in the denominator is to have the terms

$$\frac{A_1}{x+b} + \frac{A_2}{(x+b)^2} + \dots + \frac{A_n}{(x+b)^n}$$

Note that $A_2$, $A_3$ etc are constant terms.

In your case, try expressing it in the form

$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+6}$$

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Okay, I typed in your answer and got the question correct but now I'm confused: Why did you make "Dx+E" a numerator for x^2+6? ...And under what circumstances do you put 2 constants in the numerator (i.e. when would it be okay to have something like Fx+G in the numerator)? –  mathisnotmyforte Mar 4 '12 at 3:22
@mathisnotmyforte: Normally when you have a quadratic term like $x^2 + ax + b$ you put a linear term in the numerator. –  Aryabhata Mar 4 '12 at 3:26
@mathisnotmyforte: You use constants when the numerator is a power of a degree one polynomial; you use linear numerators when the denominator is a (power of a) irreducible quadratic. –  Arturo Magidin Mar 4 '12 at 3:26
When you have a quadratic term that has no real roots. –  Mike Mar 4 '12 at 3:26

There are two ways to deal with a repeated factor in the denominator, such as your $x^3$:

1. You can put a single fraction, with denominator the full repeated factor, and undetermined denominator of degree one less. In your example, since the denominator factors as $x^3(x^2+6)$, you would set up the partial fractions as $$\frac{Ax^2+Bx+C}{x^3} + \frac{Dx+E}{x^2+6}.$$

2. While the above works, for the purposes of integration you will then proceed to split up the first fraction and deal with it as $$\frac{Ax^2+Bx+C}{x^3} = \frac{Ax^2}{x^3} + \frac{Bx}{x^3} + \frac{C}{x^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3};$$ so the second way of dealing with repeated factors is to simply to that to being with: if you have a repeated factor $(x-a)^n$, you set it up with $n$ fractions, each with constant numerator: $$\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n};$$ with a repeated irreducible quadratic factor, $(x^2+ax+b)^n$, it's $$\frac{A_1x+B_1}{x^2+ax+b} + \frac{A_2x+B_2}{(x^2+ax+b)^2} + \cdots + \frac{A_nx+B_n}{(x^2+ax+b)^n}.$$ In your case, with $x^3$, you would set up that part as $$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3}.$$

Mind you: if you do the algebra right with your set-up, you will get the right answer; you'll just work a lot harder, and end up with the conclusion that $B=D=0$.

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+1: And on a side topic, I think there have been too many partial fraction questions. It might be time for an abstract duplicate. I am pretty sure you have a great answer lying around, which can, with some modifications (to cater to a general question) serve to be the parent answer. Would you be willing to do it (i.e. modify a question and your answer appropriately?) –  Aryabhata Mar 4 '12 at 3:27
Maybe I can add the general method to this one... –  Arturo Magidin Mar 4 '12 at 3:45
That looks promising! –  Aryabhata Mar 4 '12 at 7:04