# Not Understanding the Reason Behind This Type of Partial Fraction Decomposition [duplicate]

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So I have to teach partial fractions to some undergrads this week and I am looking for someone to help me understand why we solve partial fractions in cases like the following:

Suppose we have the following problem (Problem 1):

$$\int \frac{x^2-29x+5}{(x-4)^2(x^2+3)}$$

And we can decompose the integrand like the following: $$\frac{x^2-29x+5}{(x-4)^2(x^2+3)}= \frac{A}{(x-4)}+\frac{B}{(x-4)^2}+\frac{Cx+D}{(x^2+3)}$$

Contrast this with this problem (Problem 2):

$$\int \frac{3x+11}{(x-3)(x+2)}$$

We can decompose that by the following:

$$\frac{3x+11}{(x-3)(x+2)}= \frac{A}{(x-3)}+\frac{B}{(x+2)}$$

Note the second and third terms in the decomposition of Problem 1 and contrast that with both of the terms in the decomposition of Problem 2. What I do not understand is why we tack on an "x" onto our constant "C" in Problem 1. I also don't quite understand why there isn't an "x" tacked onto our constant "B" as well in Problem 1.

My question is not about how to solve either Problem 1 or Problem 2, but my question is how to explain this to undergraduates. I could go in front of my class and just explain to them the general rule to follow, but I like to provide some sort of intuition behind the rule so they aren't blindly following a step by step procedure without thinking.

## marked as duplicate by Hans Lundmark, TrueDefault, José Carlos Santos, Glorfindel, user91500Jul 31 '17 at 9:24

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• caicedoteaching.wordpress.com/2008/11/18/… – Amzoti Dec 2 '13 at 5:00
• It will be hard. For courses at a medium-low level, proving the relevant theorem may not be appropriate. Anyway, each time one does successfully a partial fractions decomposition, one has essentially proved that it is correct. The only suggestion I have is the very mild one of giving examples where the "wrong" approach fails. – André Nicolas Dec 2 '13 at 5:06
• It might be easier for you to explain what you mean if you use different symbols in partial faractions' numerators for both problems, say A,B,C,D for Problem 1 and G,H for Problem 2. – CiaPan Oct 9 '14 at 12:39

## 1 Answer

As André Nicolas says in his comment, it may be hard. I don't think one can get very far unless students have some intuition for polynomial long division, which, in my experience, is not always the case among calculus students.

If you can assume that your students understand the idea that, for polynomials $F(x)$ and $G(x),$ $$\frac{F(x)}{G(x)}=Q(x)+\frac{R(x)}{G(x)}$$ where polynomials $Q(x)$ and $R(x)$ are the quotient and remainder, the latter of which satisfies $\deg(R(x))<\deg(G(x)),$ then you can tell them that, since polynomials are straightforward to integrate, we only need to worry about integrating the second term, that is, a fraction in which the degree of the numerator is less than the degree of the denominator.

Dealing with the case where the denominator is a power of a binomial is then not so bad. In particular, we can answer your question about why there is no $x$ attached to the $B$ in your first example. The reason is that, if there were an $x$ term, it could be eliminated as in the following example: $$\frac{2x+7}{(x+3)^2}=\frac{2(x+3)+7-2\cdot3}{(x+3)^2}=\frac{2(x+3)+1}{(x+3)^2}=\frac{2}{x+3}+\frac{1}{(x+3)^2}.$$ More generally, when the denominator is a binomial raised to a power greater than $2,$ all non-constant terms in the numerator can be eliminated: \begin{aligned} \frac{4x^2+2x+7}{(x+3)^3}&=\frac{4(x+3)^2-4(6x+9)+2x+7}{(x+3)^3}=\frac{4(x+3)^2-22x-29}{(x+3)^3}\\ &=\frac{4(x+3)^2-22(x+3)+22\cdot3-29}{(x+3)^3}=\frac{4(x+3)^2-22(x+3)+37}{(x+3)^3}\\ &=\frac{4}{x+3}-\frac{22}{(x+3)^2}+\frac{37}{(x+3)^2}. \end{aligned} Essentially we're doing a change of variable from $x$ to $u=x+3,$ and using the fact that a polynomial in $x$ is a polynomial in $u$ of the same degree, to get this conclusion.

The case where the denominator is the product of different binomials is best understood by first going in the opposite direction: $$\frac{5}{x+2}+\frac{4}{x+3}=\frac{5(x+3)+4(x+2)}{(x+2)(x+3)}=\frac{9x+23}{(x+2)(x+3)}.$$ The method of partial fractions is then understood as the process of undoing this computation. This is always possible because the resulting pair of linear equations always has a unique solution. This is a consequence of the assumption that the two binomials in the denominator are different (more precisely, that they are not constant multiples of each other).

Students also need to understand that a quadratic denominator cannot always be factorized over the reals. (For example, $x^2+x+1.$) We therefore have to be able to deal with numerators up to degree $1.$ You can then show how fractions with degree $2$ and $1$ denominators get put together: $$\frac{2x+3}{x^2+x+1}+\frac{4}{x+2}=\frac{(2x+3)(x+2)+4(x^2+x+1)}{(x^2+x+1)(x+2)}=\frac{6x^2+11x+10}{(x^2+x+1)(x+2)}.$$ Partial fractions is then the method for reversing this process. It again involves solving a system of linear equations which is guaranteed to have a unique solution (assuming the denominators have no common factor). Proving this involves algebra and linear algebra knowledge that is probably beyond most first year calculus students. But this example shows that, in general, the $x$ in the numerator of the last term in your first example is needed.

Another key result needed for complete understanding is that any polynomial with real coefficients can be factorized into degree $1$ and irreducible degree $2$ factors. To prove this requires knowledge that goes quite a bit beyond first-year calculus.