Earlier in this series, I gave three different methods of showing that

Using the fact that is independent of , I’ll now give a fourth method.

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:

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:

,

where and are the positive numbers so that

,

,

so that

At this point in the calculation, we can employ the now familiar antiderivative

,

so that

.

Now it remains to simplify this fraction. To do this, we note that

,

so that

Matching coefficients, we see that

,

Since both and are positive, we see that from the last equation. Therefore,

Plugging these in, we finally conclude that

,

again matching our earlier result.

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