Answer
$$\frac{1}{{12}}\ln \left| {\frac{{x - 6}}{{x + 6}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2} - 36}}} dx \cr
& {\text{From the table of integrals in the back of the book}} \cr
& {\text{ }}\int {\frac{1}{{{u^2} - {a^2}}}} du = - \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C \cr
& \int {\frac{1}{{{x^2} - 36}}} dx = \frac{1}{{2\left( 6 \right)}}\ln \left| {\frac{{x - 6}}{{x + 6}}} \right| + C \cr
& = \frac{1}{{12}}\ln \left| {\frac{{x - 6}}{{x + 6}}} \right| + C \cr
& \cr
& {\text{Integrate by partial fractions}} \cr
& \int {\frac{1}{{{x^2} - 36}}} dx = \int {\frac{1}{{\left( {x - 6} \right)\left( {x + 6} \right)}}} dx \cr
& \frac{1}{{\left( {x - 6} \right)\left( {x + 6} \right)}} = \frac{A}{{x - 6}} + \frac{B}{{x + 6}} \cr
& 1 = A\left( {x + 6} \right) + B\left( {x - 6} \right) \cr
& x = 6 \to 1 = 12A,{\text{ }}A = \frac{1}{{12}} \cr
& x = - 6 \to 1 = - 12B,{\text{ }}B = - \frac{1}{{12}},{\text{ }} \cr
& \frac{1}{{\left( {x - 6} \right)\left( {x + 6} \right)}} = \frac{{1/12}}{{x - 6}} + \frac{{ - 1/12}}{{x + 6}} \cr
& \int {\frac{1}{{\left( {x - 6} \right)\left( {x + 6} \right)}}} dx = \int {\left( {\frac{{1/12}}{{x - 6}} + \frac{{ - 1/12}}{{x + 6}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{1}{{12}}\ln \left| {x - 6} \right| - \frac{1}{{12}}\ln \left| {x + 6} \right| + C \cr
& = \frac{1}{{12}}\ln \left| {\frac{{x - 6}}{{x + 6}}} \right| + C \cr} $$
