- #1
Prologue
- 185
- 1
In a table in two different books they both say:
[tex]\int \frac{1}{a^2-u^2}du=\frac{1}{2a} ln\left|\frac{u+a}{u-a}\right|+c[/tex]
but I am not having the same result:
[tex]\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[/tex]
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?
[tex]\int \frac{1}{a^2-u^2}du=\frac{1}{2a} ln\left|\frac{u+a}{u-a}\right|+c[/tex]
but I am not having the same result:
[tex]\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[/tex]
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?