# Integration formula

1. Jan 22, 2008

### uman

Hello all,

My textbook states the formula $\int\frac{du}{(u^2+\alpha^2)^m}=\frac{1}{2\alpha^2(m-1)}\frac{u}{(u^2+\alpha^2)^{m-1}}+\frac{2m-3}{2\alpha^2(m-1)}\int\frac{du}{(u^2+\alpha^2)^{m-1}}$ but does not provide a proof of this formula. Anyone want to show me how it's derived? I tried integration by parts, which the book gives as the method for deriving the formula, but I couldn't figure it out :-(. Any help?

2. Jan 23, 2008

### EnumaElish

You may want to try the homework section.

3. Jan 23, 2008

### uman

Why?

Also any mods can move this if necessary but it isn't homework for a class...

4. Jan 23, 2008

### EnumaElish

A "homework-type" question might get a faster response in the HW section.

5. Jan 24, 2008

### uman

You have a point. I don't know how to move it or if that's even possible. If a mod sees this please move it.

6. Jan 24, 2008

### Gib Z

An obvious trig substitution converts this integral into a power of cosine integral, which has well known recursion formulae, or otherwise, easier to derive recursion formulae than the original integral.

EDIT: The hyperbolic substitution converts it into a power of hyperbolic cosine integral, which is even easier to deal with.

Last edited: Jan 24, 2008