Earlier in this series, I gave three different methods of showing that
Since is independent of , I can substitute any convenient value of that I want without changing the value of . As shown in previous posts, substituting yields the following simplification:
if . (The cases and have already been handled earlier in this series.)
To complete the calculation, I employ the now-familiar antiderivative
Using this antiderivative and a simple substitution, I see that
This completes the fourth method of evaluating the integral , using partial fractions.