Breaking a proper rational function into pieces with simple denominators

0:00Here is a fraction with polynomials on both sides of the line, and an instruction to integrate it. Nothing you have met so far touches it. It is not a rate over the thing it is the rate of, so no substitution finds anything to substitute for. Its denominator is not a quadratic on its own, so none of the six results for quadratic denominators fits either. What you are usually handed at this point is a table of five shapes and an instruction to match your denominator against it.

0:36That table is not the method. It is the method's index -- a list of what the answers look like, set out in an order somebody chose. And almost every mark lost here is lost before you ever reach it. Two decisions come first, and both of them are easy to skip. The first decision is whether the fraction is proper. Proper means the top has the LOWER degree. Not the lower or equal degree -- the lower.

1:08So a squared term over a squared term is improper, even though nothing about it looks top heavy. An improper fraction has to be divided first. Long division turns it into a polynomial plus a proper fraction, and only the proper part goes anywhere near the shapes. Skip the division and you do not get a wrong answer. You get NO answer: the equations you write down have no solution at all, and there is nothing in them to tell you why.

1:41A student who has just been told the method reads that as an arithmetic slip and hunts for it in the wrong place for ten minutes. The second decision is whether the denominator has actually been broken up as far as it goes. Every shape you will be offered is keyed to a denominator already written as factors. If yours still has a factor hiding in it, nothing matches, and again there is no message telling you so.

2:13Now, with a proper fraction and a factorised denominator, look at the five shapes. Two distinct linear factors give two constants. One linear factor squared gives a constant over the factor and a constant over its square. Three distinct linear factors give three constants. A squared factor beside a distinct one gives three. And a linear factor beside a quadratic that will not break gives a constant over the linear one and a LINEAR numerator over the quadratic.

2:48Five rows. Ten things to hold in your head, if you hold them as five rows. They are not five facts. They are one rule, written out five times, and the rule is a count. Here is the count. Take any factor of the denominator. Say it has degree d, and say it appears m times. That factor gets m terms: one over the factor, one over its square, and so on up to its m-th power.

3:19And each of those terms gets a numerator of degree one less than the factor's. Below a linear factor, a constant. Below a quadratic, something linear. So that factor costs d times m unknowns -- exactly the degree it contributes to the denominator. Add that up over all the factors and the unknowns come to the denominator's own degree. Every time. That is the whole idea, and once you have it the five rows are consequences rather than things to remember.

3:54I ran that rule on the five factorisations the rows are keyed to. The unknowns came to the degree in all five, and the shapes it produced were the five rows, including the linear numerator on the last one. Nothing in my working read the table. It built the shapes from the factors. With a shape chosen, the rest is mechanical, and it is worth doing slowly once. Multiply both sides by the denominator. Every fraction disappears and you are left with a statement about polynomials.

4:33Now expand the right-hand side and collect the powers of the input. One equation per power. Two unknowns and two powers give two equations. Three unknowns and three powers give three. Solve them and you have the numbers. Then each piece integrates on its own, and this is the part that pays: a constant over a linear factor is a logarithm, a constant over the SQUARE of one is a reciprocal, and a linear numerator over a quadratic that will not break splits into another logarithm and an inverse tangent.

5:12Three kinds of term, out of one decomposition. I carried seven of these all the way through, with no table anywhere in the working. For each one: the shape from the counting rule, the unknowns from exact elimination over the rationals, and then an answer ASSEMBLED from the pieces rather than typed in. Every one of the seven systems had exactly one solution -- not none, and not a whole family of them.

5:44Every decomposition rebuilt to the fraction it came from, exactly, as polynomials. And every assembled answer went through the same door every answer in this series goes through: differentiate it numerically and compare with the integrand. Seven for seven. The same seven answers with their sign turned over fail that door, which is what makes the seven passes worth having. Now the step everybody performs and almost nobody justifies: comparing coefficients.

6:21Why are you allowed to say that because two expressions are equal, the number in front of each power must match? You are not, in general. Two expressions can be equal at a particular input and disagree everywhere else. You are allowed it here because the decomposition line is not an equation. It is an identity -- it holds for every value the input can take, and an identity between two polynomials really does force the coefficients to match.

6:55That distinction is worth a measurement rather than a nod. Take the simplest case and read the cleared line at twenty inputs. The numbers the elimination returns satisfy it at all twenty, and rebuild the fraction exactly. Now take a pair of numbers fitted to make the line true at one convenient input. That pair satisfies it at exactly ONE of the twenty, and it does not rebuild anything. It solved an equation. It did not find a decomposition.

7:31And the same licence buys you a shortcut, which is the reason to care about it. If an identity holds for every input, it holds at the inputs that are most convenient -- including the ones that kill most of the terms. Put a root of the denominator into the cleared line. Every term carrying that factor goes to nought, and one unknown is left standing on its own. Read it off. No system, no elimination.

8:02That is not a different method. It is the same identity, used somewhere sensible. I ran both routes against each other over six hundred cases of three distinct linear factors, with the second route written from the roots and never allowed to call the first. Six hundred agreements out of six hundred. Two mistakes are worth more than all the others, and both of them are about the shape. The first: a squared factor in the denominator needs TWO terms, one over the factor and one over its square. Not one.

8:41The instinct is that if the denominator carries the square, the term should carry the square, and one term should do. Here is what that costs. Over a hundred and fifty proper fractions with a squared linear factor beside a distinct one, the two-term shape solves every single time -- exactly one solution, a hundred and fifty out of a hundred and fifty. Drop the term over the bare factor and a hundred and forty of those hundred and fifty have no solution at all.

9:16Ten of them still solve, and those ten are the control that makes the rest mean something. The short shape works exactly where the unknown it dropped was going to be nought anyway -- at every one of the hundred and fifty cases, not merely at those ten. So it is not that one term is a harder version of two. It is that one term is a system with more equations than unknowns, and it has no answer.

9:47The second mistake is the same mistake in different clothes. A quadratic factor that will not break gets a LINEAR numerator, not a constant. That looks arbitrary until you run the count. The factor has degree two, so it costs two unknowns, so its numerator needs two coefficients. A constant has one. Measured the same way: over a hundred and twenty proper fractions with a linear factor beside a quadratic that will not break, the linear numerator solves a hundred and twenty times out of a hundred and twenty.

10:25A constant on top instead leaves a hundred and four of them with no solution. And again there is a control. Sixteen still solve, and those sixteen are exactly the cases where the input's coefficient was going to come out nought. The counting rule is not a convention. It is the difference between a system with an answer and a system without one. Now back to the first decision, because there is something wrong with how it is usually stated.

11:00One version says: proper when the top's degree is the smaller, improper otherwise. Under that version, equal degrees are improper and you divide. The other version says: divide when the top's degree is the GREATER. Under that version, equal degrees are proper and you do not. Both are in circulation, sometimes side by side, and they do not agree. I wrote them as two separate rules and ran them over twenty-four pairs of numerator and denominator. They agree on sixteen, and disagree on eight -- and the eight are exactly the equal-degree case.

11:43On those eight the difference is not a matter of taste. Follow the second version, do not divide, and all eight systems have no solution. Divide first and all eight have exactly one. Learn the second version and there is a whole family of problems you cannot start. The safe rule is the first one: if the top's degree is not strictly smaller, divide. There is one more thing the five shapes will not do for you, and it is better to meet it here than in an exam.

12:19Take a denominator that is a product of two quadratics, neither of which breaks into linear factors. Go down the five rows looking for it. It is not there. There is no row for two quadratics. That is not an oversight you can shrug off. The counting rule as I stated it works from the denominator's rational roots, and this denominator has none, so it returns nothing whatever -- not a wrong shape, no shape at all.

12:51A student told the five rows are exhaustive will conclude the question itself is faulty, and will be wrong. There are two ways out, and they land in the same place. The first is a detour. If the denominator is really a polynomial in the SQUARE of the input, substitute a new letter for that square. The two quadratics become two distinct linear factors in the new letter -- the very first row.

13:22Decompose there, in a case you can do in your sleep, and then put the square back. The two pieces are now constants over quadratics, and each is an inverse tangent. Note what the substitution was for. It was used to find the decomposition, and then undone. It was not used to do the integral. The second way is to go looking for the factors themselves. I searched a fixed grid of quadratics for one that divides the denominator exactly, and found both. The counting rule then gives four unknowns, with a linear numerator over each quadratic.

14:04The elimination returns exactly one solution -- and the two input coefficients come out nought on their own, which is a reading and not an assumption. Both answers pass the door. And across five inputs their difference takes exactly one value, and that value is nought. They are not equivalent answers. They are the same answer. One more detail, small in the working and worth real marks. A logarithm that comes out of a decomposition carries modulus bars, because the thing inside it can be negative.

14:44There is exactly one situation in which you may take them off: when you can show the argument never changes sign. The case that comes up is a substitution that leaves the logarithm of the modulus of the sine less two. The sine never reaches two. So that argument is negative everywhere, and you may turn it round and write the logarithm of two less the sine, with no bars.

15:12Measured across fifty-one angles: two less the sine stays at or above one every time, the sine less two stays at or below minus one every time, and the logarithm of the turned-round argument can be taken at all fifty-one. The control is an argument that does change sign. The input less two takes both signs over the inputs I read at, and dropping the bars there leaves a rule with nothing to say at fourteen of them.

15:45Bars come off for a reason or they stay on. To see whether any of this depends on the examples being chosen kindly, I built a grid before naming any of it. Nineteen denominators: every product of two factors drawn from a fixed list of four linear ones and two quadratics that will not break. Eighteen of the nineteen get a shape from the counting rule. The nineteenth is the product of the two quadratics, and it gets none -- the same refusal again, arriving on its own rather than because I went looking for it.

16:26Seventy-two pairs of numerator and denominator in all, and ten of them are improper -- set aside by the degree test before anything else happens. Over the sixty-two that remain, the elimination returns exactly one solution every time and never a family, and every decomposition rebuilds exactly. And every assembled answer passes the rate door. Sixty-two pass, none fails, and none goes unread. Then turn every one of those answers' signs over and run the same grid again. Nought pass and sixty-two fail -- which is what makes the sixty-two passes a measurement rather than a tally of a foregone conclusion.

17:15Step back and look at what a decomposition is actually for. It does not make the integral easier by making it shorter. It makes it easier by making it into things you already own. A constant over a linear factor is the logarithm row. A constant over a squared factor is the power row, giving a reciprocal. A linear numerator over a quadratic that will not break is two rows at once: the rate of the quadratic over itself, which is a logarithm, and what is left over, which is an inverse tangent.

17:53That last one is worth saying out loud. One term, three lines of work, two different kinds of function in the answer. Every integral this technique produces lands on a row you already have. None of them needs anything new. So what do you carry out of here? Not five shapes. One count: a factor of degree d appearing m times costs d times m unknowns, and the unknowns come to the denominator's degree.

18:26Two decisions before the shapes: is the top's degree strictly smaller, and is the denominator broken up as far as it goes. Both of them are silent when you get them wrong -- what you get back is a system with no solution, which looks exactly like an arithmetic mistake and is not one. One licence: the line is an identity, which is why you may compare coefficients, and which is also why you may substitute a root and read an unknown off in one step.

19:01And one honest limit: the shapes do not cover everything, and when they run out you substitute your way to something they do cover. The list of five is the index. What you have just built is the method.