Integration: Partial Fractions and Substitution

(New Question of the Week)

Many of the questions we answer are primarily about “how do you solve this problem?”, but at the same time ask deeper, more general questions: “How do you solve problems like this, and are there alternatives?” Today’s question is a good example of this, and raises an interesting point or two.

Can you skip the substitution?

Amia, another user we have known since the days of the old site, asked this last week:

Why do we use substitution before using partial fractions to compute the integral e^x/ (e^(2x) – 3e^x + 2)?

Is it possible to use partial fractions directly? If yes, why do we use substitution ?

All three questions have straightforward answers, which are best given by showing work for the example. This is a good reason for asking a general question with a specific case in mind. Doctor Fenton answered [emphasis mine],

Hi Amia,

It isn’t essential to substitute a new variable if you can manage the computation in your head, but it is probably easier for most students to make the substitution to clarify what needs to be done. Even if you can see that you can make the decomposition by viewing the expression as a rational expression in e^x, actually making the substitution u = e^x will show that you must decompose the fraction 1/(u^2 – 3u + 2), and not u/(u^2 – 3u + 2). You must consider the e^x in the numerator as part of the differential in the substitution du = e^x dx which arises in transforming the integral with the substitution u = e^x.

So, it is possible to carry out the decomposition and integration without explicitly making the substitution, but that approach offers opportunities to make subtle errors, while making the substitution explicitly and then carrying out the integration in the new variable is more straightforward, in my opinion.

If you have any questions, please write back and I will try to explain further.

A lot of what is taught in math classes has this basic purpose: to make a process easier for a newcomer to carry out correctly. With experience, one can skip steps and hold certain parts of a problem in one’s head; but this is risky until you are thoroughly familiar with those details you want to skip.

Amia wrote back, showing some work and making the source of the difficulty a little clearer:

Thank you Dr.

We know if we have a proper rational function, when we make decomposition, the degree of numerator will be less than the degree of denominator by 1.

If I have 1/(x – 1)(x + 3), we write a/(x – 1) + b/(x – 3).

In the above question, how can we write e^x/(e^2x – 3e^x + 2)?

Is it a/(e^x – 2) + b/(e^x – 1)?

Amia shows an understanding of the partial fractions part by stating the first step (for a different problem), but is trying to do it for the given problem directly, without either making the substitution or doing the equivalent implicitly the way Doctor Fenton indicated. So it is time to show more details, comparing the two methods and showing the error that can result:

It looks as if you did not understand the point I made in my reply.  If you are integrating \(\displaystyle \int \frac{e^x}{e^{2x} – 3e^x + 2}dx\), then making the substitution \(u=e^x\) (and \(du=e^x dx\)), the integral becomes \(\displaystyle \int \frac{1}{u^2 – 3u + 2}du\), and you find numbers a and b such that \(\displaystyle \frac{1}{(u-2)(u-1)} = \frac{a}{u-2} + \frac{b}{u-1}\) and integrate.

You seem to be asking about decomposing \(\displaystyle \frac{e^x}{(e^x-2)(e^x-1)} = \frac{c}{e^x-2} + \frac{d}{e^x-1}\).

The numerators c and d will be different from the numerators a and b found after the substitution \(u=e^x\).  If you find c and d and try to use that decomposition, you will probably not obtain the correct antiderivative.  Integrating \(\displaystyle \int \frac{c}{e^x – 2}dx\) is more complicated than integrating \(\displaystyle \int \frac{a}{u – 2}du\), since the first integral will require a more complicated substitution. \(v=e^x – 2\).

Do you understand the difference in the two approaches?

No further response came, so I will assume this explanation was sufficient. But let’s continue the discussion here.

What the alternative really looks like

First, I can finish the work using substitution. Completing the partial fraction decomposition, I get \(\displaystyle \frac{1}{(u-2)(u-1)} = \frac{1}{u-2} + \frac{-1}{u-1}\). Now I integrate \(\displaystyle \int\frac{du}{(u-2)(u-1)} = \int\left(\frac{1}{u-2} – \frac{1}{u-1}\right)du \) \(\displaystyle = \ln \left|u-2\right| – \ln\left|u-1\right| + C\). Finally, I reverse the substitution to get the answer, \(\displaystyle \ln \left|e^x – 2\right| – \ln\left|e^x – 1\right| + C\).

Second, what will the work look like if we don’t explicitly substitute? We mentally see \(e^x\) as a single entity, and \(e^x dx\) as its differential, something like (in our mind) \(\displaystyle \int \frac{d\left[ e^x\right]}{\left[ e^x\right]^{2} – 3\left[ e^x\right] + 2}\), so that we want to decompose \(\displaystyle \frac{1}{\left[ e^x\right]^{2} – 3\left[ e^x\right] + 2}\). This expands as \(\displaystyle \frac{1}{\left(\left[ e^x\right]-2\right)\left(\left[ e^x\right]-1\right)} = \frac{1}{\left[ e^x\right]-2} + \frac{-1}{\left[ e^x\right]-1}\). This leads to the exact same integral as before (though still with \(e^x dx\) in place of u everywhere), and the same answer. The only difference in the work is that we never actually wrote a u or du; but we had to see implicitly that the numerator would be 1, not \(e^x\). This required some willpower!

Now, if we missed that, and did the decomposition without doing any substitution first (as Amia wanted to do), could we in fact solve the problem? As Doctor Fenton pointed out, we will get different constants when we decompose \(\displaystyle \frac{e^x}{(e^x-2)(e^x-1)} = \frac{c}{e^x-2} + \frac{d}{e^x-1}\); in fact, I get c = 2, d = -1. So now we have to integrate \(\displaystyle \int \frac{e^x dx}{e^{2x} – 3e^x + 2} = \int \left(\frac{2}{e^x-2} – \frac{1}{e^x-1}\right)dx\).

To integrate the first term, we make the substitution \(u = e^x – 2\), so that \(du = e^x = u + 2\), and the integral becomes \(\displaystyle \int \frac{2}{e^x-2} dx = \int \frac{2}{u\left(u + 2\right)} du\). We now have to do another partial fraction decomposition! This gives us \(\displaystyle \int \left(\frac{1}{u} – \frac{1}{u+2}\right) dx = \ln\left|u\right| – \ln\left|u+2\right| = \ln\left|e^x-2\right| – \ln\left|e^x\right|\). Similarly, the second term becomes \(\displaystyle \int \left(\frac{1}{u} – \frac{1}{u+1}\right) dx = \ln\left|u\right| – \ln\left|u+1\right| = \ln\left|e^x-1\right| – \ln\left|e^x\right|\). Subtracting these, we get the same result as before.

Which would you rather do: the first method, taking our time to do the substitution and get it right; or the second method, writing less and thinking harder, but still getting it right; or the last method, using partial fractions immediately and, after a much harder slog, with two harder substitutions leading to two extra partial fractions, eventually getting the right answer again? All are valid; but the first is definitely better for beginners, and safer even for those with experience. The last gives you lots of experience, but has nothing else going for it.

So this is the answer to Amia’s question: It is possible to skip the substitution, but you do so at your own risk! In particular, in this case we do the substitution first because we see that it will lead to an integral we know how to do. In other problems that look similar, that may not happen, and we may want to try something else. Integration is an art, and often all you can do is try different ideas.

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