Answer
$$\ln \left| {x - 1} \right| - \ln \left| {x + 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{3}{{\left( {x - 1} \right)\left( {x + 2} \right)}}} dx \cr
& {\text{partial fraction decomposition}} \cr
& \frac{3}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 2}} \cr
& 3 = A\left( {x + 2} \right) + B\left( {x - 1} \right) \cr
& {\text{for }}x = 1 \cr
& 3 = A\left( {1 + 2} \right) + B\left( {1 - 1} \right) \cr
& 3 = 3A \cr
& 1 = A \cr
& {\text{for }}x = - 2 \cr
& 3 = A\left( { - 2 + 2} \right) + B\left( { - 2 - 1} \right) \cr
& 3 = - 3B \cr
& - 1 = B \cr
& {\text{substituting the values}} \cr
& \frac{A}{{x - 1}} + \frac{B}{{x + 2}} = \frac{1}{{x - 1}} - \frac{1}{{x + 2}} \cr
& \int {\frac{3}{{\left( {x - 1} \right)\left( {x + 2} \right)}}} dx = \int {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 2}}} \right)} dx \cr
& {\text{integrating}} \cr
& = \ln \left| {x - 1} \right| - \ln \left| {x + 2} \right| + C \cr} $$