# 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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