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

Question

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?

Answer 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

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

Answer 2

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 $B=D=0$.