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Let $\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$

$x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$   ...(1)

Substituting $x=1,2$, and 3 respectively in equation (1), we obtain $A=\frac{1}{2}, B=-2$, and $C=\frac{3}{2}$

$\therefore \frac{x}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}$

$\Rightarrow \int \frac{x}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\right\} d x$

$=\frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+\mathrm{C}$