# Quick Rejog of Partial Fraction Decomposition:

I don't need the problem to be solved I just need to have the decomposed equation. I have the following equation: $$$$I=\int \frac{dx}{x^2(x^2-16)}$$$$ The method that I have to use is Partial Fractions Decomposition. My question is since there is an $$x^2$$ term I have to increase in linearity which I wonder if I did correctly? $$\bbox[border:2px solid black] {$$\frac{1}{x^2(x^2-16)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4}$$}$$

• $Bx$ in the second term is redundant; you can make do with $C/x^2+\cdots$ – FearfulSymmetry Nov 25 '20 at 17:16
• @Integrand I was just confused because I thought quadratic would have that form? – EnlightenedFunky Nov 25 '20 at 17:17
• To see why it's redundant, look at what happens when you combine the first two terms: $$\frac{A}{x} + \frac{Bx+C}{x^2} = \frac{(A+B)x+C}{x^2},$$ so you don't need both $A$ and $B$. – Théophile Nov 25 '20 at 17:17

Basically you're correct, but you only need two constants, not three.

$$\frac Ax + \frac{Bx+C}{x^2} = \frac{A+B}{x} + \frac C{x^2} = \frac{A'}{x} +\frac{C}{x^2}$$
where $$A'=A+B$$.
In other words, either the $$\frac {A}x$$ term is redundant or the $$Bx$$ term is redundant.
You only increase the degree of numerator with irreducible factors (e.g. $$x^2+1$$), not repeated roots. Suppose the root $$a$$ is repeated $$r$$ times, then the decomposition has $$\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_r}{(x-a)^r}$$
One can easily observe what redundant is as follows $$\frac{1}{x^2(x^2-16)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4}$$ $$=\frac{A}{x}+\frac{B}{x}+\frac{C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4}$$ Thus either $$\frac Ax$$ or $$\frac{bx}{x^2}=\frac Bx$$ is redundant