Question:

$I = \int \frac{dx}{x^2 - a^2} = \int \frac{2a}{2a(x^2 - a^2)} dx = \frac{1}{2a} \int \frac{(x+a)-(x-a)}{(x+a)(x-a)} dx$<br>$= \frac{1}{2a} \int \left( \frac{1}{x-a} - \frac{1}{x+a} \right) dx = \frac{1}{2a} (\ln |x-a| - \ln |x+a|) + C$<br>$\therefore \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + C$