$\cfrac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}$

Let $I = \int {\cfrac{{\left( {3x - 1} \right)dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}}$

We write, $\cfrac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \cfrac{A}{{x - 1}} + \cfrac{B}{{x - 2}} + \cfrac{C}{{x - 3}}$

$\Rightarrow$ $3x - 1$
$= A\left( {x - 2} \right)\left( {x - 3} \right) + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right)$

….(i)
Putting $x = 1$ in (i),

we get
$3 - 1 = A\left( {1 - 2} \right)\left( {1 - 3} \right)$ $\Rightarrow$ $2 = A\left( { - 1} \right)\left( { - 2} \right)$ $\Rightarrow$ $A = 1$

Putting $x = 2$ in (i),

we get
$6 - 1 = B\left( {2 - 1} \right)\left( {2 - 3} \right)$ $\Rightarrow$ $5 = B\left( 1 \right)\left( { - 1} \right)$ $\Rightarrow$ $B = - 5$

Putting $x = 3$ in (i),

we get
$9 - 1 = C\left( {3 - 1} \right)\left( {3 - 2} \right)$ $\Rightarrow$ $8 = C\left( 2 \right)\left( 1 \right)$ $\Rightarrow$ $C = 4$

$\therefore$ $\cfrac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \cfrac{1}{{x - 1}} - \cfrac{5}{{x - 2}} + \cfrac{4}{{x - 3}}$

$\Rightarrow$ $I = \int {\cfrac{{3x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx}$

$= \int {\cfrac{1}{{x - 1}}dx} - 5\int {\cfrac{1}{{x - 2}}} dx + 4\int {\cfrac{{dx}}{{x - 3}}}$

$= \log \left| {x - 1} \right| - 5\log \left| {x - 2} \right| + 4\log \left| {x - 3} \right| + C$