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
From Sign of Quotient of Factors of Difference of Squares:


 * $x^2 > a^2 \implies \dfrac {x - a} {x + a} > 0$

Then: