Answer
$$\frac{1}{8}\ln \left| {\frac{{x - 1}}{{x + 7}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2} + 6x - 7}}} dx \cr
& {\text{completing the square for the denominator}} \cr
& {x^2} + 6x + {\left( {\frac{6}{2}} \right)^2} - {\left( {\frac{6}{2}} \right)^2} - 7 \cr
& \left( {{x^2} + 6x + 9} \right) - 9 - 7 \cr
& {\left( {x + 3} \right)^2} - 16 \cr
& \cr
& \int {\frac{1}{{{x^2} + 6x - 7}}} dx = \int {\frac{1}{{{{\left( {x + 3} \right)}^2} - 16}}} dx \cr
& {\text{make }}u = x + 3,\,\,\,\,{\text{then}}\,\,\,\,\,\,du = dx \cr
& \int {\frac{1}{{{{\left( {x + 3} \right)}^2} - 16}}} dx = \int {\frac{1}{{{u^2} - 16}}} du \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 70}} \cr
& \left( {70} \right):\,\,\,\,\,\,\int {\frac{{du}}{{{u^2} - {a^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C \cr
& \int {\frac{1}{{{u^2} - 16}}} du = \frac{1}{{2\left( 4 \right)}}\ln \left| {\frac{{u - 4}}{{u + 4}}} \right| + C \cr
& \cr
& {\text{write in terms of }}x{\text{; replace }}x + 3 {\text{ for }}u \cr
& = \frac{1}{8}\ln \left| {\frac{{x + 3 - 4}}{{x + 3 + 4}}} \right| + C \cr
& = \frac{1}{8}\ln \left| {\frac{{x - 1}}{{x + 7}}} \right| + C \cr} $$
