The one problem I missed, 30 years ago, on my final exam in calculus

It’s been said that we often remember our failures more than our successes. In this instance, the adage rings true, because I can still remember, clear as a bell, the one problem that I got wrong on my high school calculus final that I took 30 years ago. Here it is:

\displaystyle \int (x^2+1)^2 dx

I tried every u-substitution under the sun, with no luck. I tried u = x^2+1. However, du would be equal to 2x \, dx, and there was no extra x in the integrand.

I believe I tried every crazy, unorthodox u-substitution possible given the time constraints of the exam: u = \sqrt{x}, u = \sqrt{x^2+1}, u = 1/x. Nothing worked.

We had learned trigonometric substitutions in my class, and so I also tried those. I started with x = \tan u, so that x^2 + 1 = \tan^2 x + 1 = \sec^2 x. This looked promising. However, dx = \sec^2 u \, du, so the integral became \displaystyle \int \sec^4 u \, du. From there, I was stuck. (Now that I’m older, I know that the logical train actually goes in the reverse direction than what I attempted as a student.)

I wasn’t taught integration by parts in this first course in calculus, so I didn’t even know to try it. Had I known this technique, I probably would’ve broken through my conceptual barrier to finally get the right answer. (In other words, integration by parts will yield the correct answer, but it’s a lot of work!) But I didn’t know about it then, and so I get to tell the story now.

Exasperated, I turned in my exam when time was called, and I asked my teacher how this integral was supposed to be solved.

Easy, she told me: just square out the inside:

\displaystyle \int (x^2+1)^2 dx = \displaystyle \int (x^4 + 2x^2 + 1) \, dx = \displaystyle \frac{x^5}{5} + \frac{2x^3}{3} + x + C

At the time, I was unbelievably annoyed at myself. Now, I love telling this anecdote to my students as I relate to their own frustrations as they practice the art of integration.

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