Answer
$$ - \frac{1}{6}\ln \left| {x - 1} \right| + \frac{1}{{15}}\ln \left| {x + 2} \right| + \frac{1}{{10}}\ln \left| {x - 3} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}} \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 2}} + \frac{C}{{x - 3}} \cr
& {\text{Multiplying by }}\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right){\text{ yields}} \cr
& 1 = A\left( {x + 2} \right)\left( {x - 3} \right) + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right)\left( {x + 2} \right) \cr
& {\text{substituting }}x = 1 \cr
& 1 = A\left( 3 \right)\left( { - 2} \right) + B\left( 0 \right) + C\left( 0 \right) \cr
& A = - \frac{1}{6} \cr
& {\text{substituting }}x = - 2 \cr
& 1 = A\left( 0 \right) + B\left( { - 3} \right)\left( { - 5} \right) + C\left( 0 \right) \cr
& B = \frac{1}{{15}} \cr
& {\text{substituting }}x = 3 \cr
& 1 = A\left( 0 \right) + B\left( 0 \right) + C\left( 2 \right)\left( 5 \right) \cr
& C = \frac{1}{{10}} \cr
& \cr
& {\text{Then}}{\text{, the integrand can be written as}} \cr
& \frac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}} = \frac{{ - 1/6}}{{x - 1}} + \frac{{1/15}}{{x + 2}} + \frac{{1/10}}{{x - 3}} \cr
& \int {\frac{{dx}}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}} = \int {\left( {\frac{{ - 1/6}}{{x - 1}} + \frac{{1/15}}{{x + 2}} + \frac{{1/10}}{{x - 3}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = - \frac{1}{6}\ln \left| {x - 1} \right| + \frac{1}{{15}}\ln \left| {x + 2} \right| + \frac{1}{{10}}\ln \left| {x - 3} \right| + C \cr} $$
