I'm asking about definite integrals that can effortlessly be found numerically by high schoolers using software. For example,

$$\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$$

This link shows the numerical integration done by software.

This Reddit comment substantiates that the exact solution is knotty, but not why:

The extent of math that this involves (beyond standard integration techniques usually taught in Calc II, just applied on a large scale), is a significant bit of Complex Analysis (the residue theorem, etc.). In general, everything in his derivation should at least be understandable had you taken Calc I-III and Complex Analysis.

How can I explain to someone who is just getting started in calculus why it's so hard to find the exact solution by symbolic integration, when it's so trivial to get a numerical answer using software?

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    $\begingroup$ Related: math.stackexchange.com/q/3242391/18398 $\endgroup$
    – JRN
    Commented May 28, 2019 at 8:16
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    $\begingroup$ I had a hard time understanding the original version of the question, and other people's comments suggest that I wasn't the only one. Only after re-reading it several times and clicking on the links did I realize what the OP was actually asking. I've rewritten the question to try to make it clearer, while hopefully not losing the OP's intent. $\endgroup$
    – user507
    Commented May 30, 2019 at 13:15
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    $\begingroup$ I tried plugging this equation into a TI84 calculator (this model is the "standard" the the teachers in my HS use), and I quickly got a 'divide by 0' error. I then adjusted to integrate from -1 to 0 and then 0 to 1 and got a sum of 8.372205547 which is different than the reported result on the linked page, after the 5th digit. Can I assume that the software 'numerical integration' uses a different algorithm between the 2? $\endgroup$ Commented May 30, 2019 at 22:00
  • $\begingroup$ Just as, for a five year old, learning addition, subtraction, multiplication, division may be incredibly difficult for most, they can easily be trained to use a calculator to "perform" the operations. That doesn't mean they understand how the operations work, just as being able to strictly use CAS for approximating integrals, numerically, doesn't mean students understand what, in fact, integration is. $\endgroup$
    – amWhy
    Commented May 31, 2019 at 17:10
  • $\begingroup$ Tell students at the beginning of the course that many antiderivatives can not be solved analytically. This is not the only thing like this. For instance roots of most 5th+ polynomials are not analytically soluble. Or some simple equations involving log or sin with small polynomials. When they get to diffyscrews, many of those will not be soluble, even some deceptively simple ones. (Yes, in this case, perhaps it is analytically soluble using an array of tricks and complex analysis...but the point remains. Actually gives some motivation for learning more tricks and complex analysis!) $\endgroup$
    – guest
    Commented Jan 6, 2020 at 14:35

7 Answers 7


It might be fun to have your students pretend that the only functions they know are sums of monomials $cx^n$ where $n\in{\bf Z}$, and in particular, play like they know nothing about the function $f(x)=\ln(x)$. You see where this is going: we can easily graph $f(x)=x^n$ between $x=1$ and $x=2$ for any $n$, and find an antiderivative with the one annoying exception of $f(x)=x^{-1}$.

So, what is done in some developments of calculus? We define a new function $g(x)=\int_{1}^x \frac{1}{t}\, dt$ precisely so that we can have an antiderivative for the situation where we were previously stuck, and give $g(x)$ the special name $\ln(x).$

The point is that integration, unlike differentiation, takes one outside the function type of the integrand, e.g., the derivatives of rational functions are rational functions, but their antiderivatives might not be.

Wrap up the lesson with the xkcd comic found here



You should expect numerical integration to be "easier" [1] than symbolic integration because it is answering a fundamentally weaker question. That is, symbolic integration, if you can do it, gives you an infinitely precise answer. Numerical integration gives you an answer which is approximate, ideally with some bound on the amount of error.

It is like the difference between $\sqrt{2}$ and $1.414 \pm 0.0003$.

[1]: You might not find numerical integration quite as easy if you were to do it by hand. If you level the playing field and allow the computer for symbolic integration, it can often be pretty easy too.

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    $\begingroup$ Also, numerical integration is answering a weaker question because the limits of integration have to be specified. $\endgroup$ Commented Jan 6, 2020 at 14:13

Because integrals are not something to be "solved" by default, but a tool of mathematical expression in their own right.

Can you express $\sqrt{2}$ as something else? Probably nothing simpler. You can approximate it, but that's about it. Likewise, why should you expect to be able to express some arbitrary integral, say

$$\int_{0}^{1} e^{\sin x}\ dx$$

as something simple? But of course, generating arbitrarily precise approximations to either is trivial, because it basically amounts in both cases to cutting off the limiting definitions at a finite number of steps, or finding some more efficient procedure that is equivalent thereto, and hence is systematic. In a sense, numerical integration is already there within the definition of integration itself, as a limit of the Riemann sum.

The real problem this relates to is a) not emphasizing enough the distinction between a mathematical object and a representation of that object, and b) thinking that there is "one true and correct" representation. When you "solve" an integral, what you're really doing is trying to create a representation of the value (or function, if it's an indefinite integral) it denotes, with something that is either not itself an integral, or doesn't have to be one, at least (already one can suggest that the natural logarithm is, in effect, the first "non-elementary" integral, as it is the first one to take one outside the field operations of the real numbers, but we call it elementary because tradition).

But not only may such representations not exist, but even if they do, they might not be the best representations to use in a given situation. For example, this indefinite integral

$$\int \sqrt[3]{\tan x}\ dx$$

famously starred in a YouTube video by a maths professor going under the handle "blackpenredpen", because while it can be "solved" symbolically with "elementary functions", the expression is a long, crazy, awful mess, so bad that if you encountered it in a problem somewhere, I would actually think it preferable to simply leave it as an integral, with a small note that the integral has an elementary symbolic representation, but it is far too cumbersome to use.

Indeed, this extends to the whole idea that you should always "simplify" everything. Actually - no. What you should really seek is to put things into a form that makes their intent the most clear, and/or that makes them the most pleasing aesthetically. In many cases, that may involve simplification, but not all. Same for reducing fractions, etc. - I've seen tons of times when that an important pattern in some sequence of fractions would be seen far more clearly if one did not reduce them. And likewise, I would not call the expression coming from the integral immediately above a "simplification" - more like a "complication" to me. Yes, it gets rid of the integral, but it's very difficult to decipher any useful meaning from it. Moreover, one can in fact also argue it doesn't even do that: tangent, and other trigonometric functions, ultimately need some kind of integral-like process to define them, so you're in effect "massaging" some integrals into other integrals, that are worse and more verbose and not at all clean. Same with my very first example. There are ways to express it using various fascinating and obscure mathematical functions, but in terms of sheer elegance, I think simply leaving it as the integral actually is the best justice. There's no need, as you in effect have hinted at, for such expressions for doing calculations, when powerful numerical integrators exist already.

Once you stop seeing integrals as things needing to be "solved", any more than $\sqrt{2}$ "needs" some kind of "solving" to be done to it when you encounter it, then questions like these reveal themselves to be rather unnecessary, or their answer in fact quite trivial: Integrals are really a much, *much* more powerful mathematical operation than the ones you have had on hand before you are introduced to them, and they should be thought of as opening up whole new vistas of expressibility of things you could not express before, and that is the meaning of "why they're hard". Reducing them to things for "solving" is to, in a way, insult this power, and suggest they are an obstacle, instead of a facility. Integration is not hard. Trying to tame the power it represents, and shrink it down to what you call "familiar", is.

And so terms like "Exact forms", "closed forms", etc. are things that I think should be banished, too, for the same reason. There is nothing "inexact" about any of the above expressions. We should instead talk of various kinds of forms - representation by integrals, by arithmetic functions, by trig/log functions, by codified integral-defined functions ($\mathrm{erf}$, etc.), by infinite sums, and so forth and not get hung up on any one in particular as "better" or "worse" but to get a good feel for the pros, cons and utility of each. Maths is really a language to express, not a puzzle to crack. Puzzles are what you express in maths, just as you can express them in human language. And ideally, just as we should strive for clarity in expressing ourselves in our human languages, we should do the same for our mathematical language, as well.

  • $\begingroup$ Why did you delete this answer? It seems excellent to me. $\endgroup$ Commented Jan 16, 2020 at 15:04
  • $\begingroup$ @Chris Cunningham : Not sure now, hmm ... $\endgroup$ Commented Jan 17, 2020 at 4:04
  • $\begingroup$ Wonderful answer! $\endgroup$ Commented Mar 8 at 14:44

When students learn integration, it's usually the first time they've ever encountered a type of mathematical calculation for which there are techniques, but no general algorithm that always succeeds. A good way of explaining why differentiation is rule-based, but integration is not, is to make the analogy with discrete, finite sums.

If you want to evaluate the sum $f(n)=1+2+3+\ldots+n$, you can easily find the result for any given $n$, but it's harder to find a general formula for $f(n)$. OK, so next show the student how to visualize triangles of dots and use geometrical reasoning to find that the answer is $n(n+1)/2$.

But you can now demonstrate that going the opposite way is just a straightforward calculation of $f(n)-f(n-1)$, with a simplification to show that the result is $n$. This rule-based calculation is like a derivative.

So in your example of using software to numerically evaluate a definite integral, what the software is doing is essentially like finding that $1+2+3=6$. That's straightforward and rule-based. But that doesn't mean that we can find the general formula $n(n+1)/2$ in the same rule-based way.

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    $\begingroup$ Ah man, this is a great insight I was looking for! I always attributed the phenomenon to the arbitrariness of what most are willing to call a 'closed form' (what if erf is a named/special function for you but you don't consider exp simple or important enough to give a name?) or treated it as a byproduct of that system, but differentiation comprises two evaluations (the usual system of writing being biased so as to make $f(x+\delta)$ (or the difference $f(y)-f(x)$) usually 'simplifiable' or analyzable) and discarding terms that are $o(\delta)$, both of which tend to be amenable to $\endgroup$ Commented Jun 2, 2019 at 15:07
  • $\begingroup$ convenient writing. Whereas (discrete) integration involved adding an indefinite number of terms (or for normal integration, perhaps you can think of a hyperreal number $N$), and it's a miracle when that sum, as a function of $N$, happens to match one that one counts as familiar. Thank you for finding words for the intuition! $\endgroup$ Commented Jun 2, 2019 at 15:11

I would point out that integration is the reverse of differentiation, a pretty straightforward process. But often times the reverse of an easy algorithm is surprisingly hard!

  • Solving a linear system of equations vs. plugging a value into linear equations

  • Factoring vs. expanding (polynomial and prime factoring)

  • Finding the inverse of a function at a point vs. computing the function at a point ($f^{-1}(a)$ vs. $f(a)$)

  • some P vs. NP problems

Even on a basic level, most people find division harder than multiplication. Radicals are harder than exponents (natural exponents at least). Differentiation and integration are no different.


This reminds me of one of Euler's papers, where he evaluates (not the definite integral, but indefinite integral) $$\int\frac{dz}{(3\pm z^2)\sqrt[3]{1\pm 3z^2}}$$ using an amazing variety of substitutions and other tricks, among other things the formula $\arctan(x)-\arctan(y) = \arctan\left(\frac{x-y}{1+xy}\right)$. (In fact, he evaluates it two different ways!)

I think you can feel perfectly free to say that one the greatest mathematicians of all time struggled with various unusual integrals, even of the type which are actually doable with elementary antiderivatives, and then published papers about them. Maybe we don't do that as much today. And that apparently he also thought it wasn't evident which ones are easy and which ones aren't.

But far from being a discouragement, this should be instead freedom for a student. Use creativity, and see what you get! Use all the tricks you have access to now, acquire more later, and keep trying. And these are good thoughts for life; know what you've got, have a growth learning mindset to get more, and persevere.

  • $\begingroup$ We rarely tell students that integration is actually guess-work. We can do the standard integrals now because we have a list of good guesses. $\endgroup$
    – Jessica B
    Commented May 29, 2019 at 6:20
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    $\begingroup$ @JessicaB, it can be taken beyond guesswork. $\endgroup$ Commented May 29, 2019 at 7:30
  • $\begingroup$ True. Though note the part of the Wikipedia page connecting this to the undecidable 'does this expression equal zero' problem. In any case, my point was more the discovery than actual integration of any particular integrand. $\endgroup$
    – kcrisman
    Commented May 30, 2019 at 4:27
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    $\begingroup$ @Peter Taylor, the Risch algorithm might “fail” by determining that the anti-derivative is not an elementary function. Then guesswork resumes...perhaps by using elliptic functions, or perhaps the definite integral can be evaluated by contour integration, etc. $\endgroup$
    – user52817
    Commented Jun 1, 2019 at 18:31

One way of thinking about it (although I'm not sure whether it would be helpful for the student of interest to you):

What are you doing when you 'graph' the function? At each point on the $x$-axis, you calculate the value of the function, and plot that on the $y$-axis. That is something that can be done at each value. In practice, we start by knowing this is a 'reasonably nice' function (ie mostly continuous), so you can get away with finding the value at finitely many points and either stopping (on a computer screen) or joining the dots (in your head).

Now, what are you doing when you 'integrate' the function? Finding the area under the graph. Well, sort of. Finding the area under the graph can be done. An integral does exist (ignoring issues about measure etc). That's a point students don't always appreciate. There is an integral, and it is some function. The problem is how to write down what that function is. When a student asks 'can I integrate this?' what they really mean is 'can I write the integral of this function in a reasonably nice formula using functions I am familiar with?'

I think phrased that way, it becomes clear why integration is 'harder' than graphing, which should be sufficient explanation for most 17-year-olds.

(Now I've written it out, I can see that this is similar to the point made by user52817 .)


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