My first question on math.SE asked for different ways to evaluate . This post will discuss my original method for evaluating , the method in my favorite answer from the math.SE question, and why the two are equivalent (even though they do not appear to be at first glance).
One way to evaluate is to multiply the integrand by (or, as I tell my students, by in the right way). This yields
The numerator is now the derivative of the denominator, and so a substitution with produces an integral of the form . Thus we have
My favorite answer from my math.SE question (although there are multiple good ones) is the one by Derek Jennings. Rewrite as and multiply by (again, by in the right way). With this, we have
Partial fractions decomposition on the last integral yields
Why are these equal? They certainly don’t look like the same answer. They are, though, even up to the same constant C. Starting with the second answer, multiply the fraction inside the logarithm by (once again, by in the right way). This produces
the same expression as in the first answer.