Answer
$$\frac{1}{8}\ln \left| {\frac{{x - 7}}{{x + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} - 6x - 7}}} \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& \frac{1}{{{x^2} - 6x - 7}} = \frac{1}{{\left( {x - 7} \right)\left( {x + 1} \right)}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{\left( {x - 7} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 7}} + \frac{B}{{x + 1}} \cr
& {\text{Multiplying the equation by }}\left( {x - 7} \right)\left( {x + 1} \right){\text{, we have}} \cr
& {\text{1}} = A\left( {x + 1} \right) + B\left( {x - 7} \right) \cr
& {\text{if we set }}x = 7, \cr
& {\text{1}} = A\left( {7 + 1} \right) + B\left( 0 \right) \cr
& A = \frac{1}{8} \cr
& {\text{if we set }}x = - 1, \cr
& {\text{1}} = A\left( 0 \right) + B\left( { - 1 - 7} \right) \cr
& B = - \frac{1}{8} \cr
& {\text{then}} \cr
& \frac{1}{{\left( {x - 7} \right)\left( {x + 1} \right)}} = \frac{{1/8}}{{x - 7}} + \frac{{ - 1/8}}{{x + 1}} \cr
& \cr
& \int {\frac{{dx}}{{{x^2} - 6x - 7}}} = \int {\left( {\frac{{1/8}}{{x - 7}} + \frac{{ - 1/8}}{{x + 1}}} \right)dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{8}\int {\frac{1}{{x - 7}}dx} - \frac{1}{8}\int {\frac{1}{{x + 1}}} dx \cr
& {\text{integrating}} \cr
& = \frac{1}{8}\ln \left| {x - 7} \right| - \frac{1}{8}\left| {x + 1} \right| + C \cr
& = \frac{1}{8}\ln \left| {\frac{{x - 7}}{{x + 1}}} \right| + C \cr} $$
