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

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

My answer:

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

What am I doing wrong?

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    [link](http://www.wolframalpha.com/input/?i=%28x%5E4%2B2%29%2F%28x%5E5%2B6x%5E3%29) and press show steps near "Partial fraction expansion"2012-03-04

2 Answers 2

1

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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    When you have a quadratic term that has no real roots.2012-03-04
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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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    That looks promising!2012-03-04