U-Substitution? U Gotta Be Kidding

- Calculus

This week in class we've been learning the interesting things for which area's under a curve provide an answer. And whats finding a definite integral but taking a couple antiderivatives and finding their difference? Nothing to it, right? Wrong!

Let me explain the situation. Currently, I've spent most of a semester working with the differential branch of calculus. I have a fair understanding of what derivatives are, what they represent. Having spent the first weeks of the semester using limits to find the slopes of tangent lines, I certainly have an appreciation for how powerful they are. But, most importantly, I've spent enough time taking derivatives that I've memorized their basic rules and can identify them on sight.

So now we are being introduced to integrals in class. That is, now we are thinking in terms of finding a function \(g(x)\) who's derivative would equal \(f(x)\); now the game is about trying to reverse the process. Like every new concept it's a stretch at first. Knowing this, the earliest definite integrals we find involve functions for which we can easily identify the derivative rule that needs to be reversed.

Then today, this happens:

\[ \begin{equation} \int\frac{e^u}{(1-e^u)^2}du. \end{equation}\]

When I was assigned this problem today, somewhere, surely, a sad trumpet sounded. I froze up, my notes from class mere chicken scratches that would surely have earned me negative points on a test for causing my professor excessive unwanted confusion. In short, I failed miserably. The reason is that I rushed to solve the problem, before understanding the problem. "Well, this is u-substition", I thought, "I just need to let u equal, uhh... something."

Having had some time to digest this problem, I felt it best if I codify a strategy for myself, one that I can explain and justify. Being able to write about and explain math concepts helps force me to think about them and truely understand them. Or at least I hope.

So, enough exposition, it's time to solve the problem at hand and in doing so, identify some concepts that will help with solving other, similar problems.

The u in u-substitution is defined by you

We have a little bit of an issue with our problem, it's already in terms of \(u\), so substituting with \(u\) doesn't make any sense. That's perfectly fine, though. We don't have to use \(u\) for the subsitution, the reason is that in the final solution, the substitution is undone. Alternatively, in the case of this problem, the \(u\) could be replaced by an \(x\) in the original function and then, at the end, converted back to \(u\). The important concept is that we are making a substitution to help us solve a problem. So long as as the substitutions are undone before declaring a solution final, we can make whatever manipulations help us to solve the problem.

Deciding on what to substitute

Let's look at the problem again:

\[ \begin{equation} \int\frac{e^u}{(1-e^u)^2}du \end{equation}\]

Notice anything interesting about the two factors, \(e^u\) and \(1 - e^u\)? If you guessed that one is the derivative of the other, you just hit the jackpot. Now, lets see if that plays out.

Let \(x = 1 - e^u\). Ok, good start! Now, noticing that the integral has a differential in it, \(du\), let's find a differential equation using this new equation we've created.

\[ \begin{equation} x = 1 - e^u \end{equation}\]

\[ \begin{equation} \begin{split} \frac{dx}{du} & = \frac{d}{du}(1-e^u) \\ \frac{dx}{du} & = -e^u \\ dx & = -e^{u}\ du \end{split} \end{equation}\]

Now we have two pluggable pieces with which to substitute, \( x = 1 - e^u \) and \( dx = -e^u\ du \). First up, lets plug in the \( x \) and see where that leads.

\[ \begin{equation} \int\frac{e^u}{(x)^2}du \end{equation}\]

Cleaning this up just slightly gives the following,

\[ \begin{equation} \int\frac{1}{(x)^2}\ e^u\ du \end{equation}\]

Hanging out there at the end is an expression that looks suspiciously like the one we earlier determined to be \( dx \), \( dx = -e^u\ du \). In fact, if we just move that negative to the other side it would look exactly the same: \( -dx = e^u\ du \). Therefore, we can plug \( -dx \) in place of \( e^u\ du \). Since the negative in front is really a -1, a constant, it can be moved out front of the integral:

\[ \begin{equation} -\int\frac{1}{(x)^2}\ dx \end{equation}\]

Manipulating the integrand algebraically will produce an integral that consists of a simple power function.

\[ \begin{equation} -\int x^{-2}\ dx \end{equation}\]

Just a simple power function? Now this is something I know how to integrate. Adding 1 to the power and then multiplying by its inverse gives:

\[ \begin{equation} -\left(\frac{-1}{1}\right)x^{-1} + C \end{equation}\]

The negatives out front cancel out, reducing \( x \)'s coefficient to 1. Now through with the need of the negative exponent to take advantage of the power rule, we can move it back down to the denominator.

\[ \begin{equation} \frac{1}{x} + C \end{equation}\]

All that remains is to undo the substitution that was done at the beginning of our work, that is, replace \( x \) with \( 1 - e^u \). Doing so, we arive at the solution.

\[ \begin{equation} \frac{1}{1 - e^u} + C \end{equation}\]


It pays to have a good strategy for solving problems. Good strategies start by identifying what the problem is that is being solved. In the case of u-substitution, the problem is generally needing to apply the Fundamental Theory of Calculus, but having an integral that doesn't match the form. Applying u-substitution allows us to find the areas under complex curves that wasn't previously possible, at least with a calculus 1 student's skill set. I'm sure this substitution rabbit-hole goes even further, I'll just have to get some sleep and find out tomorrow in class.