Some thoughts on exam preparation

It feels like the exams are an age away, but believe me, the coming weeks will fly by and before you know it the exams will be upon you. There are things that you can do now, which are going to make the period before the exams, and the exams themselves be a lot less stressful than they would be otherwise.

Part of the issue with MAM1000 is that it’s a big course, which covers a lot of material. It will have been many months since you studied limits by the time you get to the exam, so keeping your hand in with all the topics covered will make life much easier for you.

Of course you need to do the tutorials each week, but I recommend, starting now, going through a couple of old tuts each week. It shouldn’t take too long as you have done them before, and the more times you do them, the faster they will go. By the time you’ve run through them all two or three times, you will be able to sit down and race through an entire tut in under half an hour. This practice will make the process of solving these problems feel much more automatic.

The second hint is to find a few new problems on each topic and practicing them each week. Try and find a couple of new integration by substitution questions, a couple of limits questions, a couple of derivatives that you’ve never tried before, just so that not only are you speeding up the general process of running through the questions, but you are also getting used to seeing new problems.

All of this will put you in a great position when, two weeks before the exams you are already feeling very comfortable with material. Two weeks before the exams should really be a time of tuning, not a time of serious training – that should have happened already.

As before, when you are studying:

  • Put your phone away
  • Make sure that you are in a place where you are not going to be disturbed
  • Set yourself a series of questions to do, and if you can’t get them, don’t look up the answers until you’ve tried again the next day, and the day after, and still not been able to do them. If when you look at the answer you still don’t understand it, then come see me, or send me an email


Right, onto some more integration techniques – I know that you’re loving these!


Integration by partial fractions


Warning: The following is going to feel a little overwhelming and abstract until we look at examples, but soon enough it’s going to be really clear what’s going on. Read the following and try and follow along. In the next post we will deal with lots of examples where the theory will be put into practice and the process will become clear!


Would you know how to solve the integral:




How about the integral


\int \frac{2}{x-1}-\frac{1}{x+2}dx?


I imagine that the answer to the first question is no, and I hope that the answer to the second is yes.

Well, actually they are the same thing!




We’re going to find a general way to factor a fraction of polynomials (a polynomial in the numerator and denominator) into a form which can easily be integrated by simple integration techniques.

Actually, we’re not going to be solving many integrations in this lecture, we’re just going to be converting complicated fractions into so called partial fractions. There are going to be a number of steps in this process, which we will list now, and go over in detail in the coming examples. The steps of our process are:

  1.  Convert the expression into a proper rational form
  2.  Factor the denominator into linear terms of the form (x-a) (which may be raised to integer powers: (x-a)^m) and irreducible quadratic terms (ax^2+bx+c) (which may also be raised to integer powers (ax^2+bx+cb)^n).
  3. Convert the expression into a sum of terms, either with a power of a linear term, or a power of a quadratic irreducible in the denominator.
  4. Take the irreducible degree two terms and complete the square to get: ((x-r)^2+b^2)^n.
  5. solve the integrals in the sum one by one, either directly or by integration by substitution.


It truly doesn’t matter if this doesn’t make sense yet – I’m just giving you a roadmap, which you will learn to read as we go along.

Let’s take first things first. We want to convert the integrand into proper rational form. First we need to define what we mean be the degree of a polynomial. A polynomial is of degree n if the highest power of x in the polynomial is of the form x^n:




is a degree 6 polynomial.


We are looking at fractions of polynomials. Proper rational form means that we never have a polynomial in the numerator of higher, or the same degree as the polynomial in the denominator.


If we have a fraction of two polynomials P(x) and Q(x)




where degree(P(x))\ge degree(Q(x)), we will want to convert this into a form:


\frac{P(x)}{Q(x)}\rightarrow R(x)+\frac{S(x)}{Q(x)}


Where R(x) is some polynomial and degree(S(x))<degree(Q(x)). Note that this is equivalent to writing something with integers in the form:




We perform this conversion into proper ration form by long division of polynomials. For instance, make sure that you can show that:




Here we have taken a fraction of polynomials on the right which is not in proper rational form, and converted it into proper rational form.

It is easy to come up with examples to try yourself. Simply think of two random polynomials where one has a higher, or equal degree as the other, and find their ratio, with the higher power polynomial in the numerator. Then convert this into proper rational form by long division of polynomials. I’m going to assume in most of what follows that we’ve already completed this step and so we already have an expression in proper rational form.

If you have any problems with polynomial long division, take a look here.

OK, now that we have our expression in proper rational form, the term of the form R(x) is trivial to integrate, so we won’t even worry about it – it’s just a polynomial, and polynomials are easy to integrate, right?

For the other term (which is now a fraction of polynomials with a higher degree in the denominator than the numerator) we are going to show that we can always convert this fraction into a sum of terms of the form:








Where A,B,C are constants to be determined a,b,c,d,e,f are constants which will come directly from our original expression, and i is a positive integer.


The way we work out what all of these constants are is first of all by factorising the denominator. Let’s look at an example. Here’s an expression already in proper rational form and we factorise the denominator into linear terms in x (in this example we can do this, but we can’t always get it into exactly this form…we’ll see what to do later if that’s the case).




Now we will write a sum of terms with the factors in the denominator on their own:




It is now our job to figure out what A and B are. We do this by combining the fractions:




We see that in the numerator we have collected the terms linear in x and those with no factors of x


We are trying to solve for A and B. so now we equate our final expression with our original expression:


\frac{5x-7}{(x-2)(x-5)}= \frac{x(A+B)-(5A+2B)}{(x-2)(x-5)}


For this to be true 5=(A+B) and 7=(5A+2B). We can solve this easily to get A=-1 and B=6. Now we plug it back in to the expression where we had separated the fractions:




Check to make sure you see that this is true. Now, when faced with the seemingly very complicated integral


\int\frac{5x-7}{(x-2)(x-5)}dx we can see that this is the same as doing:


\int \frac{-1}{x-2}+\frac{6}{x-5}dx


Which you should certainly be able to do.


Ok, this was quite a simple example. What did we do?


  1. Factorised the denominator into linear terms
  2. Wrote a sum of terms with one of linear factors in each one and with unknown constants in the numerator
  3. Recombined the terms into a single fraction
  4. Matched powers of x and constant terms and solved for the unknowns
  5. Wrote the final expression in the form of the sum of the simple factors, now with the unknowns known!


This is going to be the case when we can factor the numerator into a product of distinct linear terms, but that’s not always the case. Sometimes you will end up with an expression of the form:




In this case our sum of terms will be:




If we have n powers of a particular factor in the denominator, then this will contribute n terms, with all powers of the factor from 1 to n. As an extreme example:


\frac{3x^3+4x-2}{(x+2)^2(x-4)^3} would be written as:




You would then combine all of these factors together and solve for A,B,C,D and E by matching powers of x in the numerator.


Some polynomials can’t be factored into terms of the form (x+a). For instance if we have ax^2+bx+c where b^2-4ac<0 we can’t solve this in the form (x+a_1)(x+a_2) (at least not yet). In these cases we will keep them in the unfactored form, as a quadratic (known as an irreducible quadratic, as we can’t reduce it to a product of linear terms). So, when we take our original expression (in proper rational form) we will want to, as stated before, factor it into linear terms in x and quadratic terms in x which can’t be factored. We follow the same rules for these quadratic terms when we split up our expression, but rather than having a constant in the numerator, we will have a term linear in x. So, for instance if we had:




We will write this as:




You see that the linear term (x-2) has a constant in the numerator, and the quadratic term has an expression linear in x in the numerator.

We will then combine the terms together and match the coefficients in the numerator for each power of x.

If we have a quadratic to some power, then again we will include multiple factors for it:




Wow, this sounds like quite a process. Step away from the computer, breath, and come back in ten minutes….


Back? Good! I promise it’s not that complicated, but it feels a bit overwhelming the first time.

How do we integrate these terms with unfactorisable quadratics? Well, instead of factorising them, we complete the square and the perform a substitution. For instance if we had:




We would write:




Now substitute u=x+1:


\int \frac{3(u-1)+2}{u^2+1}du=\int \frac{3u}{u^2+1}-\frac{1}{u^2+1} du


And this can be solved with trig substitutions using u=\tan p.

So, we’ve talked about polynomials which can be factored into a product of linear terms in x and irreducible quadratics. What about polynomials which can’t be factored in this way? Well, as we will learn later in the course. All polynomials, however complicated, can be factored into this form, so once you know how to deal with products of this form you can take the integral of any ratio of polynomials, however complicated. That’s a pretty strong statement!!

OK, I’m going to break for now as it might all seem a bit much for one post, and I’ll write another post with lots of examples so you can see what’s going on for real.


 A couple of interesting bibliographic diversions…

I briefly mentioned a mathematician by the name of Terry Tao when I was talking about the twin primes conjecture. He is a truly remarkable researcher and there is a nice writeup of this very impressive human being here.

There’s also a nice article about one of the most interesting living mathematicians, John Conway here.


How clear is this post?