Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 2

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {x^2 - a^2} = \frac 1 {2 a} \ln \left({\frac {x - a} {x + a} }\right) + C$

where $x^2 > a^2$.

Proof
First note that:

Then:

Also see

 * Primitive of Reciprocal of $a^2 - x^2$: Logarithm Form: Proof 2