Answer
$$\frac{1}{6}\ln \left| {\frac{{x + 1}}{{x + 7}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} + 8x + 7}}} \cr
& {\text{Factor the integrand}} \cr
& \frac{1}{{{x^2} + 8x + 7}} = \frac{1}{{\left( {x + 7} \right)\left( {x + 1} \right)}} \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{1}{{\left( {x + 7} \right)\left( {x + 1} \right)}} = \frac{A}{{x + 7}} + \frac{B}{{x + 1}} \cr
& {\text{Multiply the expression by }}\left( {x + 7} \right)\left( {x + 1} \right) \cr
& 1 = A\left( {x + 1} \right) + B\left( {x + 7} \right) \cr
& \cr
& {\text{if we set }}x = - 7 \cr
& 1 = A\left( { - 7 + 1} \right) + B\left( { - 7 + 7} \right),{\text{ }}A = - \frac{1}{6} \cr
& {\text{if we set }}x = - 1 \cr
& 1 = A\left( { - 1 + 1} \right) + B\left( { - 1 + 7} \right),{\text{ then }}B = \frac{1}{6} \cr
& {\text{Then}}{\text{,}} \cr
& \frac{A}{{x + 7}} + \frac{B}{{x + 1}} = \frac{{ - 1/6}}{{x + 7}} + \frac{{1/6}}{{x + 1}} \cr
& \int {\frac{{dx}}{{{x^2} + 8x + 7}}} = \int {\left( {\frac{{ - 1/6}}{{x + 7}} + \frac{{1/6}}{{x + 1}}} \right)} dx \cr
& = \frac{1}{6}\int {\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 7}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{6}\left( {\ln \left| {x + 1} \right| - \ln \left| {x + 7} \right|} \right) + C \cr
& = \frac{1}{6}\ln \left| {\frac{{x + 1}}{{x + 7}}} \right| + C \cr} $$
