Answer
$$\frac{1}{6}\ln \left| {\frac{{x - 3}}{{x + 3}}} \right| + C\,$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2} - 9}}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{1}{{{x^2} - 9}}} dx = \int {\frac{1}{{{x^2} - {3^2}}}} dx \cr
& {\text{The integrand has a expression in the form }}{u^2} - {a^2}{\text{}}{\text{}} \cr
& {\text{Use formula 70}} \cr
& \left( {70} \right):\,\,\,\,\int {\frac{{du}}{{{u^2} - {a^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C\,\,\, \cr
& {\text{let }}u = x,\,\,\,a = 3 \cr
& \int {\frac{1}{{{x^2} - {3^2}}}} dx = \frac{1}{{2\left( 3 \right)}}\ln \left| {\frac{{x - 3}}{{x + 3}}} \right| + C\,\,\, \cr
& {\text{simplifying}} \cr
& \int {\frac{1}{{{x^2} - 9}}} dx = \frac{1}{6}\ln \left| {\frac{{x - 3}}{{x + 3}}} \right| + C\, \cr} $$
