Integrate the rational function: $\frac{x}{(x-1)(x-2)(x-3)}$

Let $\frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.
Multiplying both sides by $(x-1)(x-2)(x-3)$,we get:
$x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.
To find $A$,put $x=1$:
$1 = A(1-2)(1-3) \Rightarrow 1 = A(-1)(-2) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2}$.
To find $B$,put $x=2$:
$2 = B(2-1)(2-3) \Rightarrow 2 = B(1)(-1) \Rightarrow 2 = -B \Rightarrow B = -2$.
To find $C$,put $x=3$:
$3 = C(3-1)(3-2) \Rightarrow 3 = C(2)(1) \Rightarrow 3 = 2C \Rightarrow C = \frac{3}{2}$.
Thus,$\frac{x}{(x-1)(x-2)(x-3)} = \frac{1}{2(x-1)} - \frac{2}{x-2} + \frac{3}{2(x-3)}$.
Integrating both sides:
$\int \frac{x}{(x-1)(x-2)(x-3)} dx = \frac{1}{2} \int \frac{1}{x-1} dx - 2 \int \frac{1}{x-2} dx + \frac{3}{2} \int \frac{1}{x-3} dx$.
$= \frac{1}{2} \log |x-1| - 2 \log |x-2| + \frac{3}{2} \log |x-3| + K$.