# Integral from Table

1. Jul 30, 2008

### Prologue

In a table in two different books they both say:
$$\int \frac{1}{a^2-u^2}du=\frac{1}{2a} ln\left|\frac{u+a}{u-a}\right|+c$$

but I am not having the same result:

$$\int \frac{1}{a^2-u^2}du=\frac{1}{2a} \left(\int \frac{1}{a-u}du + \int \frac{1}{a+u}du\right) = \frac{1}{2a} \left(-ln\left| a-u \right|\right)+ln\left| a+u \right|+c = \frac{1}{2a} ln\left|\frac{a+u}{a-u}\right|+c$$

Obviously the u+a at top is interchangeable but I am not seeing an explanation for the bottom (u-a vs. a-u) being different. Is it just an absolute value thing?

2. Jul 30, 2008

### nicksauce

Well |a-u| = |u-a|, so yes I would say its just an absolute value thing.