Rrerku

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
$$
I=intdfracdxxsqrtx^2-1.
$$
Some of my students gave the following answer.

Let $t=1/x$ then $dx=-1/t^2dt$, so we get
$$
I=intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfrac-dtsqrt1-t^2=-arcsinleft(tright)+C=-arcsinleft(frac1xright)+C.
$$

Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfractrighttsqrt1-t^2.
$$
Then we end up with the answer
$$
intdfracdxxsqrtx^2-1=begincases
-arcsinleft(dfrac1xright)+C & textfor x>1,\
arcsinleft(dfrac1xright)+C & textfor x<-1.
endcases
$$
In your teaching practice, how would you usually proceed?

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:
$$lim_xtopminfty fracxsqrtx^2+1 = pm 1$$

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

$$lim_xtopminfty fracxcdot 1/xsqrt(x^2+1)cdot 1/x^2 = lim_xtopminfty frac1sqrt1+1/x^2=1$$

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 $xto -infty$, 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 $frac1xsqrtx^2-1$ is definitely negative there.

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!)

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.)

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.