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I have recently learned about the use of Partial Fraction Decomposition (P.F.D) in integration.

I want to know whether one has to memorize the decompositions for various fractions or is there some logic behind them from which we can quickly figure 'them' out ( 'them' referring to knowing whether the numerator should be a constant, linear or quadratic polynomial just by seeing the particular fraction; etc.).

Consider this fraction $$\dfrac{x^4 +x +1}{x^3 \cdot (x^2 +9)^2}$$

I've read that its decomposition is: $$\dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x^3} + \dfrac {Dx+E}{x^2 +9} + \dfrac{Fx+G}{{(x^2 +9)}^2}$$

I'm this the numerator of 3 terms is a constant but then suddenly the rest two terms have linear expressions as their numerator. So, if I got even one of the terms wrong my calculations would go wrong too! And that is just another example of the several different fractions we get.

So, Is their some technique or logic that can be used for figuring out the P.F.D or do we actually need to memorize all of that?

HeWhoMustBeNamed
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    The numerator always has degree one less than the denominator (it represents the possible remainders on division by the denominator). – lulu Oct 24 '17 at 11:12
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    The terms with denominators that are not all squared or higher powers (i.e. those of the form $x^n+ax^{n-1}+\dotsb+c$) have numerators of degree at most one less. A denominator of the form $(x^n+ax^{n-1}+\dotsb+c)^m$ has degree at most $n-1$, as explained in this answer. – Chappers Oct 24 '17 at 11:26
  • @lulu, why doesn't it have a degree, say, two less than that of the denominator? – HeWhoMustBeNamed Oct 24 '17 at 12:07
  • Because that would not cover every possible remainder. – lulu Oct 24 '17 at 12:08
  • @lulu, sorry I didn't understand that (because I don't understand what you mean by "it represents the possible remainders on division by the denominator"). – HeWhoMustBeNamed Oct 24 '17 at 12:09
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    The division rule for polynomials says that if you are dividing $f(x)$ by $g(x)$ the form is $f(x)=g(x)\times q(x)+r(x)$. Where $q(x),r(x)$ are polynomials and $\deg r(x)<\deg g(x)$. Thus even if your integrand has a simple, irreducible, denominator, you need to allow a numerator of degree $1$ less. – lulu Oct 24 '17 at 12:11
  • @lulu, alright I think I understand- it means that we want a general form that can take all forms of permissible numerators, right? Also, can you suggest someplace I can read more about PFD's (including things like what you just explained which my book somehow skips over). – HeWhoMustBeNamed Oct 24 '17 at 12:16
  • I'd search on line first. Most references tend to treat it as a algebraic device without a lot of discussion. To be sure, if the denominator is complicated then the method doesn't tend to lead to a very pleasant integral. – lulu Oct 24 '17 at 12:23
  • @lulu sorry for the late reply but I just want to say thank you for answering my doubts. The answers below explained the 'what' of my question but not the 'why' but you explained that because of which I understand it better now. So... Thank you for the help! ;-) – HeWhoMustBeNamed Oct 25 '17 at 11:17
  • @lulu Hey I still have a doubt- in the question I gave the numerator of the last fraction doesn't have a degree one less than that of the numerator so how do we take care of all the possible remainders in this case? – HeWhoMustBeNamed Oct 25 '17 at 11:23
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    Let's just look at powers of $x$. Instead of writing $\frac Ax +\frac B{x^2}+\frac C{x^3}$ you could add those fractions to get $\frac {Ax^2+Bx+C}{x^3}$. That is the form my approach leads you to. Thus what you wrote is equivalent to what I said...which you prefer is a matter of taste. – lulu Oct 25 '17 at 11:26
  • @lulu, Ok, I get that now. Thanks for explaining (and for the patience too)! – HeWhoMustBeNamed Oct 25 '17 at 11:43

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The logic is just the general theorem on partial fractions decomposition, valid over any field:

Let $f$ and $g$ be non-zero polynomials over a field $K$. If $$g =\prod_{i=1}^n p_i^{r_i}$$ is a decomposition of $g$ as a product of (distinct) irreducible polynomials There are (unique) polynomials $b$ and $a_{ij}$ with $\deg a_{ij} < \deg p_i$ ($1\le i\le n$, $1\le j\le r_i$) such that $$\frac fg = b + \frac{a_{11}}{p_1}+\dots+\frac{a_{1r_1}}{p_1^{r_1}}+\dots + \frac{a_{n1}}{p_n}+\dots+\frac{a_{nr_n}}{p_n^{r_n}}$$ and $b$ is the quotient of the Euclidean division of $f$ by $g$. In particular, if $\deg f<\deg g$, $b=0$.

When the field is $\mathbf R$, the irreducible polynomials are linear polynomials (degree $1$) and quadratic polynomials with complex roots, hence the degree condition for numerators implies these numerators are constants and linear polynomials respectively.

Bernard
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You don't have to do anything in life, except do taxes and die.

But yeah, if you want to be any good at integration, you should remember how to decompose a rational function into partial fractions. The rule is that the numerator is always a polynomial of degree one less than the denominator before applying further powers to it. That is, if the original fraction has $(P(x))^n$ in the denominator, then this means the decomposed partial fractions will have $$\frac{P_1(x)}{P(x)} + \frac{P_2(x)}{P^2(x)} + \cdots + \frac{P_n(x)}{P^n(x)}$$ where each of the polynomials $P_i$ has a degree of $\deg(P)-1$.

So, since the original fraction has $(x^2+9)^2$ in the denominator, this will yield two fractions in the decomposition: $$\frac{P_1(x)}{(x^2+9)}\text{ and } \frac{P_2(x)}{(x^2+9)^2}$$

where both $P_1$ and $P_2$ are of degree $1$ (one less than $x^2+9$).

G Tony Jacobs
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5xum
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  • But the last partial fraction has degree 1 which is 3 less than that of its denominator. – HeWhoMustBeNamed Oct 24 '17 at 11:20
  • @MrReality It's one less than the denominator $x^2 + 9$ before squaring, you're right. I edited my answer. Also, if you downvoted me, please remove the downvote. Just because my answer was a little unclear that doesn't make it wrong. – 5xum Oct 24 '17 at 11:24
  • Can you please expand on that in your answer? And no, I did not downvote your answer. – HeWhoMustBeNamed Oct 24 '17 at 11:42
  • @MrReality I did expand on my answer. In fact, I completely rewrote the last part of my answer. Is anything still unclear? – 5xum Oct 24 '17 at 11:43
  • OK, so we have to look at the original fraction's denominator, identify all the multiplicands, group the same ones together, Remember beforehand their (simpler) decompositions and then break up the fraction such that we have a numerator with a degree one less than that of the corresponding denominator's basic unit. Is that right? Can I take it as a general algorithm which works in All cases? – HeWhoMustBeNamed Oct 24 '17 at 11:52
  • @MrReality That sounds about right, yeah. – 5xum Oct 24 '17 at 11:54
  • Dying is at your own discretion. Paying taxes is not. –  Oct 24 '17 at 11:54
  • In the list of fractions, shouldn’t those denominators be raised to powers? – G Tony Jacobs Oct 24 '17 at 11:55
  • @5xum, thank you :-), I wonder why I wasn't taught about this (Was it supposed to be obvious, eh? ) Also do you know of a good source (maybe the book you all first read since you all knew about it) where I can read more about PFD's (which has explanations like you provided).. – HeWhoMustBeNamed Oct 24 '17 at 12:00
  • @MrReality I don't know why you weren't taught this. I was taught in the first year of calculus... – 5xum Oct 24 '17 at 12:02
  • I find, as a calculus teacher, that most texts present the general decomposition without much explanation. Making sense of it is left to the student, or to the instructor to explain the logic behind it. – G Tony Jacobs Oct 24 '17 at 12:04
  • @5xum can you suggest a good source (ebook, website, whatever) then? – HeWhoMustBeNamed Oct 24 '17 at 12:04
  • @MrReality Hm. I learnt it from a pretty good textbook but it's in Slovenian... Probably any good textbook on introductory calculus should do the trick. – 5xum Oct 24 '17 at 12:05
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    @YvesDaoust Well, dying is sort of half-way at your own discretion. Your account closes at some point, it's just that you have an additional opt-out available at any time. – 5xum Oct 24 '17 at 12:11