I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?

Let's consider the following example.

Find the indefinite integral $$ I=\int\dfrac{dx}{x\sqrt{x^{2}-1}}. $$ Some of my students gave the following answer.

Let $t=1/x$ then $dx=-1/t^{2}dt$, so we get $$ I=\int\dfrac{-1/t^{2}dt}{\frac{1}{t}\sqrt{\frac{1}{t^{2}}-1}}=\int\dfrac{-dt}{\sqrt{1-t^{2}}}=-\arcsin\left(t\right)+C=-\arcsin\left(\frac{1}{x}\right)+C. $$

Sometimes, I accept this answer since it gives a quick general antiderivative. However, the problem here is that we should write $$ \int\dfrac{-1/t^{2}dt}{\frac{1}{t}\sqrt{\frac{1}{t^{2}}-1}}=\int\dfrac{-\left|t\right|dt}{t\sqrt{1-t^{2}}}. $$ Then we end up with the answer $$ \int\dfrac{dx}{x\sqrt{x^{2}-1}}=\begin{cases} -\arcsin\left(\dfrac{1}{x}\right)+C & \text{for }x>1,\\ \arcsin\left(\dfrac{1}{x}\right)+C & \text{for }x<-1. \end{cases} $$ In your teaching practice, how would you usually proceed?

PS. We may encounter the same issue in many other problems. For example, find $\int\sqrt{1-x^{2}}dx$. Then if we let $x=\sin\left(t\right)$ then $\sqrt{1-\sin^{2}\left(t\right)}$ should be $\left|\cos\left(t\right)\right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.

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    $\begingroup$ Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue. $\endgroup$ Commented Mar 25, 2019 at 16:44
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    $\begingroup$ The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain. $\endgroup$
    – Javier
    Commented Mar 25, 2019 at 19:06
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    $\begingroup$ By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :) $\endgroup$
    – kcrisman
    Commented Mar 26, 2019 at 1:49
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    $\begingroup$ Very nice question. Inattention to this detail is indeed a shortcoming of many texts and many of my own lectures. Probably adding a condition ($x>1$ or $x<-1$) to focus attention on one case is the smart solution to not overwhelm students and yet be true to detail. Some texts are more careful than others... $\endgroup$ Commented Mar 27, 2019 at 1:44
  • $\begingroup$ Obligatory xkcd $\endgroup$
    – Wildcard
    Commented Mar 27, 2019 at 10:36

6 Answers 6


I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.

It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)


This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.

Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here: $$\lim_{x\to\pm\infty} \frac{x}{\sqrt{x^2+1}} = \pm 1$$

The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:

$$\lim_{x\to\pm\infty} \frac{x\cdot 1/x}{\sqrt{(x^2+1)\cdot 1/x^2}} = \lim_{x\to\pm\infty} \frac{1}{\sqrt{1+1/x^2}}=1$$

But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.

In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $x\to -\infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)

So in your case, I would go more naive as well. Do a very rough sketch of $\arcsin(1/x)$ (you can basically do this by drawing $-\arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $\frac{1}{x\sqrt{x^2-1}}$ is definitely negative there.

enter image description here

Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.

Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)

  • $\begingroup$ And perhaps, one should be asking why in the first place students are "so used to manipulation of this kind". In my experience, this is a product of bad mathematical pedagogy, because the syllabus is often designed around 'how to do X and Y' rather than 'why should X or Y be true?'... The first question a good student should ask is "Why?". Your example is a good one to illustrate the error here, which is the failure to understand precisely what $\sqrt{x}$ is defined as. $\endgroup$
    – user21820
    Commented Mar 27, 2019 at 8:30
  • $\begingroup$ Well, my point of view is that, in one semester, I can't fix 12 years of students being taught math is only manipulation. (Not that every student has only had this experience, but enough will have had it that I need to make it a part of my planning.) So I do my best to get them to start thinking along that road at every opportunity. $\endgroup$
    – kcrisman
    Commented Mar 27, 2019 at 13:19
  • $\begingroup$ Yup I understand that. =) $\endgroup$
    – user21820
    Commented Mar 27, 2019 at 13:22

You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.

In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!

The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.


To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.

I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.

As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.


This sort of example is not artificial and needs more careful treatment than it often receives. The inherent difficulty of the example is compounded by the tendency to discuss primitives without discussing their domains.

The badly named "indefinite integral" of a function is really an incompletely defined primitive of a function. By "incompletely defined" I mean that when specifying a primitive of a function one should also specify its domain. When this is done, the issue described in the question becomes clearer. Frequently the domain is not mentioned, and the indefinite integral, which is in any case an equivalence class of objects (also confusing for students), is treated as a single well-defined entity whose precise meaning is determined by the context in which it is used. This is like the common practice of physicists, and works well when done well (it usefully avoids too much fussing about what for beginning students are perhaps marginal issues), but generates great confusion when not handled with a certain care (attention to the confusion it possibly generates).

The issue arises when solving ordinary differential equations. For example, consider the ODE whose solutions include the examples given in the question. Which case is relevant depends on the initial conditions. One approach when teaching this is to always state initial value problems rather than asking for the "general solution". At any rate, it is probably not helpful to students to ask for the "general solution" when there is an ambiguity like that in the question, unless of course one wants to focus attention on this ambiguity.

I once taught calculus with a professor who insisted on never writing an integral without limits. This practice eliminates the ambiguity described in the question. It worked quite well. While I'm not as inflexible about it as he was, I try to minimize writing indefinite integrals in contexts where confusion could arise.

One also runs into essentially the same issue when teaching multivariable calculus, where it is related to real physical phenomena. Whether a vector field is conservative or not depends on the domain in which it is considered. Classical examples are vector fields such as $\left(\tfrac{-y}{x^{2} + y^{2}}, \frac{x}{x^{2}+ y^{2}}\right)$ on $\mathbb{R}^{2}$ or $\tfrac{x}{|x|^{3}}$ on $\mathbb{R}^{3}$, which are singular at the origin, and admit a potential in a ball not containing the origin, but do not in a ball containing the origin, a fact related to the Gauss-Coulomb law in electrostatics. Failure to condition the definition of "conservative" or "exact" on the domain leads to confusion for students who are trying to understand how it can be that an irrotational vector field is not a potential.


It is interesting to note that if we instead write the antiderivative as $I=-\arctan\left(\frac{1}{\sqrt{x^2-1}}\right)$, then this form is valid for both $x<1$ and $x>1$. In other words, using this arctan representation, we avoid a need for a piecewise representation for the antiderivative.

This problem is well suited for formative assessment. I would give full credit for an answer of the form $I=-\arcsin\left(\frac{1}{x}\right)$, and then use this as a launching point for a discussion of these more involved issues that you have uncovered, depending on the mathematical maturity of the student.

  • $\begingroup$ A piecewise representation is still technically necessary, since this antiderivative has two independent parameters. $\quad$ Its simplest form is $\operatorname{arcsec}|x|.$ $\endgroup$
    – ryang
    Commented Mar 15, 2023 at 18:12

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