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I don't need the problem to be solved I just need to have the decomposed equation. I have the following equation: \begin{equation} I=\int \frac{dx}{x^2(x^2-16)}\end{equation} 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? $$\bbox[border:2px solid black] {\begin{equation}\frac{1}{x^2(x^2-16)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4} \end{equation}}$$

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    $\begingroup$ $Bx$ in the second term is redundant; you can make do with $C/x^2+\cdots$ $\endgroup$ – FearfulSymmetry Nov 25 '20 at 17:16
  • $\begingroup$ @Integrand I was just confused because I thought quadratic would have that form? $\endgroup$ – EnlightenedFunky Nov 25 '20 at 17:17
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    $\begingroup$ 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 $B$. $\endgroup$ – Théophile Nov 25 '20 at 17:17
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Basically you're correct, but you only need two constants, not three.

As mentioned in the comments,

$$ \frac Ax + \frac{Bx+C}{x^2} = \frac{A+B}{x} + \frac C{x^2} = \frac{A'}{x} +\frac{C}{x^2}$$

where $A'=A+B$.

In other words, either the $\frac {A}x$ term is redundant or the $Bx$ term is redundant.

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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 $$\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_r}{(x-a)^r}$$

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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

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    $\begingroup$ Re-welcome....hiiii :-) $\endgroup$ – Sebastiano Nov 25 '20 at 23:34

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