Answer
$$\frac{1}{8}\ln \left| {\frac{{x - 5}}{{x + 3}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} - 2x - 15}}} \cr
& {\text{completing the square}} \cr
& \int {\frac{{dx}}{{{x^2} - 2x - 15}}} = \int {\frac{{dx}}{{{x^2} - 2x + 1 - 16}}} \cr
& = \int {\frac{{dx}}{{\left( {{x^2} - 2x + 1} \right) - 16}}} \cr
& = \int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2} - 16}}} \cr
& u = x - 1,{\text{ }}du = dx \cr
& = \int {\frac{{du}}{{{u^2} - 16}}} \cr
& = \int {\frac{{du}}{{\left( {u + 4} \right)\left( {u - 4} \right)}}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{\left( {u + 4} \right)\left( {u - 4} \right)}} = \frac{A}{{u + 4}} + \frac{B}{{u - 4}} \cr
& 1 = A\left( {u - 4} \right) + B\left( {u + 4} \right) \cr
& u = - 4 \to A = - \frac{1}{8} \cr
& u = 4 \to B = \frac{1}{8} \cr
& \frac{1}{{\left( {u + 4} \right)\left( {u - 4} \right)}} = - \frac{{1/8}}{{u + 4}} + \frac{{1/8}}{{u - 4}} \cr
& = \int {\frac{{du}}{{\left( {u + 4} \right)\left( {u - 4} \right)}}} = \int {\left( { - \frac{{1/8}}{{u + 4}} + \frac{{1/8}}{{u - 4}}} \right)} du \cr
& = \frac{1}{8}\ln \left| {u - 4} \right| - \frac{1}{8}\ln \left| {u + 4} \right| + C \cr
& = \frac{1}{8}\ln \left| {\frac{{u - 4}}{{u + 4}}} \right| + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{8}\ln \left| {\frac{{x - 1 - 4}}{{x - 1 + 4}}} \right| + C \cr
& = \frac{1}{8}\ln \left| {\frac{{x - 5}}{{x + 3}}} \right| + C \cr} $$
