We have
\(\frac 1{x^2 -a^2} \)
\(= \frac 1{(x-a)(x+a)}\)
\(= \frac 12 \left[\frac{(x+a)-(x-a)}{(x-a)(x+a)}\right]\)
\(= \frac 1{2a}\left[\frac 1{x-a}-\frac 1{x+a}\right]\)
Therefore,
\(\int \frac{dx}{x^2 -a^2}\)
\(= \frac 1{2a}\left[ \int \frac{dx}{x-a} - \int \frac{dx}{x+a}\right]\)
\(= \frac 1{2a} [\log|(x-a)| - \log |(x+a)|]+C\)
\(= \frac 1{2a} \log\left|\frac {x-a}{x + a} + C\right|\)
We have
\(\int \frac{dx}{x^2 -16}= \int \frac{dx}{x^2 -4^2} \)
\(= \frac 18\log\left|\frac {x -4}{x + 4}\right| +C\)
