Answer
$$\frac{1}{3}\ln \left| {\frac{{x - 2}}{{x - 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} - x - 2}}} \cr
& {\text{completing the square}} \cr
& \int {\frac{{dx}}{{{x^2} - x - 2}}} = \int {\frac{{dx}}{{{x^2} - x + \frac{1}{4} - \frac{9}{4}}}} \cr
& = \int {\frac{{dx}}{{\left( {{x^2} - x + \frac{1}{4}} \right) - \frac{9}{4}}}} \cr
& = \int {\frac{{dx}}{{{{\left( {x - \frac{1}{2}} \right)}^2} - \frac{9}{4}}}} \cr
& u = x - \frac{1}{2},{\text{ }}du = dx \cr
& = \int {\frac{{du}}{{{u^2} - \frac{9}{4}}}} \cr
& = \int {\frac{{du}}{{\left( {u - 3/2} \right)\left( {u + 3/2} \right)}}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{\left( {u - 3/2} \right)\left( {u + 3/2} \right)}} = \frac{A}{{u - 3/2}} + \frac{B}{{u + 3/2}} \cr
& 1 = A\left( {u + 3/2} \right) + B\left( {u - 3/2} \right) \cr
& u = \frac{3}{2} \to A = 3 \cr
& u = - \frac{3}{2} \to B = - \frac{1}{3} \cr
& \frac{A}{{u - 3/2}} + \frac{B}{{u + 3/2}} = \frac{{1/3}}{{u - 3/2}} + \frac{{ - 1/3}}{{u + 3/2}} \cr
& = \int {\frac{{dx}}{{{x^2} - x - 2}}} = \int {\left( {\frac{{1/3}}{{u - 3/2}} + \frac{{ - 1/3}}{{u + 3/2}}} \right)} du \cr
& = \frac{1}{3}\ln \left| {u - 3/2} \right| - \frac{1}{3}\ln \left| {u + 3/2} \right| + C \cr
& = \frac{1}{3}\ln \left| {\frac{{u - 3/2}}{{u + 3/2}}} \right| + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{3}\ln \left| {\frac{{x - 1/2 - 3/2}}{{x - 1/2 + 3/2}}} \right| + C \cr
& = \frac{1}{3}\ln \left| {\frac{{x - 2}}{{x - 1}}} \right| + C \cr} $$
