# How to obtain the rational function integration recursive formula

1. Jan 21, 2014

### mr_sparxx

I've been dealing with several integrals involving rational functions. I have encountered myself arriving to an integral that requires the application of the following recursive formula:

$\int\frac{1}{(u^2+α^ 2)^m} \, du= \frac{u}{2 α^ 2 (m-1)(u^2+α^ 2)^{m-1}}+\frac{2m-3}{2 α^2 (m-1)}\int\frac{1}{(u^2+α^ 2)^{m-1}} \, du$

as stated in Apostol's Calculus.

However, I am curious about the demonstration of this formula. Apostol states in his book that it is obtained by integrating by parts, but I don't see how... does anybody have any ideas?

Thanks!

Bibliography
Calculus, Volume 1, One-variable calculus, with an introduction to linear algebra, (1967) Wiley, ISBN 0-536-00005-0, ISBN 978-0-471-00005-1

2. Jan 21, 2014

### tiny-tim

hi mr_sparxx!

put Im = $\int\frac{1}{(u^2+α^2)^m} \, du$

then Im-1 = $\int\frac{u^2+α^2}{(u^2+α^2)^m} \, du$

= $\int\frac{u^2}{(u^2+α^2)^m} \, du$ + α2Im

= (something)$\left[\frac{u}{(u^2+α^2)^{m-1}}\right]$ + (something)Im-1 + α2Im

carry on from there

3. Jan 21, 2014

### mr_sparxx

Awesome! :)
Thanks tiny-Tim (specially for leaving some of the fun for me) ;P