$$\begin{align*}\int\frac{1}{x^2-a^2}dx&=\frac{1}{2a}\int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)d{x}\\&=\frac{1}{2a}\left(\int\frac{1}{x-a}d{x}-\int\frac{1}{x+a}d{x}\right)\\&=\boxed{\frac{1}{2}\left(\ln|{x-a}|-\ln|{x+a}|\right)+C}\\&=\boxed{\ln\left|{\frac{x-a}{x+a}}\right|+C}\end{align*}$$