A Reduction Formula When using a reduction formula to solve an integration problem, we apply some rule to rewrite the integral in terms of another integral which is a little bit simpler. We may have to rewrite that integral in terms of another integral, and so on for n steps, but we eventually reach an answer. For example, to compute: � (ln x)n dx we repeat the integration � by parts from the previous example n − 1 times, until we’re just calculating (ln x) dx. For our ﬁrst step we use: u = (ln x)n v=x u� = n(ln x)n−1 1 x v � = 1. Then: � n (ln x) dx n = x(ln x) − n = x(ln x)n − n � � 1 (ln x)n−1 x dx x (ln x)n−1 dx So, if: � Fn (x) = (ln x)n dx then we’ve just shown that: Fn (x) = x(ln x)n − nFn−1 (x). This is an example of a reduction formula; by applying � the formula repeatedly � we can write down what Fn (x) is in terms of F1 (x) = ln xdx or F0 (x) = 1 dx. We illustrate the use of a reduction formula by applying this one to the preceding two examples. We start by computing F0 (x) and F1 (x): � F0 (x) = (ln x)0 dx = x + c F1 (x) = x(ln x)1 − 1F0 (x) (use reduction formula) = x ln x − x + c (Example 1) F2 (x) = x(ln x)2 − 2F1 (x) (use reduction formula) = x(ln x)2 − 2(x ln x − x) + c = x(ln x)2 − 2x ln x + 2x + c This is how reduction formulas work in general. 1 (Example 2.)