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Why is it useful to write:

$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$

and not:

$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$

when decomposing into partial fractions?

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6  
One short answer is because the second one doesn't work. –  T. Bongers Jul 16 at 0:35
    
Why does not it work? –  saadtaame Jul 16 at 0:37
    
Just try the normal process, and you'll find it's not possible to put numbers in the numerators there. –  T. Bongers Jul 16 at 0:39
    
Alternatively, it does work if you take the numerator of the $(x+1)^-2$ term to be linear. Either way, this case requires three parameters in the partial fraction decomposition. –  Semiclassical Jul 16 at 0:42
    
The idea of partial fractions is to decompose into simple terms with numerators that are constants. The second one will work if you allow for numerator to be linear in $x$ but that can be easily shown to be equal to the first method. Another way to see that is to think of $(x+1)^2$ as having multiplicity two and writing it at $(x+1)(x+1+\epsilon)$ and then using the normal method of partial fractions taking $\epsilon=0$ at the very end. –  suresh Jul 16 at 0:43

2 Answers 2

For example, try decomposing $\frac{1}{(x+1)^2(x-1)}$ using the second form: \begin{align} \frac{1}{(x+1)^2(x-1)} &= \frac{A}{(x+1)^2} + \frac{B}{x-1} \\ \frac{1}{(x+1)^2(x-1)} &= \frac{A(x-1) + B(x+1)^2}{(x+1)^2(x-1)}. \end{align} Then $B$ must be zero by looking at the $x^2$ term, and then $A$ must be zero as well. So as the commenter pointed out, it just doesn't work. But using the first form we get $$\frac{1}{(x+1)^2(x-1)} = -\frac{1}{2 (x+1)^2}-\frac{1}{4 (x+1)}+\frac{1}{4 (x-1)}.$$

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Suppose you have a rational function $r(x) = \frac{p(x)}{q(x)}$ with $p(x)$ and $q(x)$ polynomials and the degree of $q(x)$ is larger than the degree of $p(x)$. You can examine its behavior near the zeroes of $q(x)$. If $q(y) = 0$ then you can expand $r(x)$ around $x= y$ as:

$$r(x) = \frac{A_n}{(x-y)^n} + \frac{A_{n-1}}{(x-y)^{n-1}} +\cdots$$

where $n$ is the multiplicity of the zero at $x=y$ of $q(x)$. Suppose that to approximate $r(x)$ globally, you add up the expansion for each zero of $r(x)$ and you keep only the singular terms. Then that would not just be an approximation to $r(x)$, it would be the same as $r(x)$, because the difference between $r(x)$ and the sum of all the singular parts opf all the expansions obviously has no singularities left. Therefore that difference is a polynomial. However, this polynomial tends to zero at infinity because $r(x)$ tends to zero at infinity and all the singular terms we have subracted also tend to zero. This means that the polynomial is in fact identical to zero.

So, you see that each term of the partial fraction expansion comes from the expansion around the singularities of the rational function and you need to keep all the singular terms, not just the most dominant ones with the largest negative power.

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