Solution:

$\frac{3 x-1}{(x-4)(x+1)(x-1)}=\frac{A}{x-4}+\frac{B}{x+1}+\frac{C}{x-1}$

$\therefore 3 x-1=A(x+1)(x-1)+B(x-4)(x-1)+C(x-4)(x+1)$

$Put x=4$

$3(4)-1=A(4+1)(4-1)$

$\therefore 11=15 A$

$\therefore A=\frac{11}{15}$

$Put x=-1$

$3(-1)-1=B(-1-4)(-1-1)$

$\therefore-4=B(-5)(-2)$

$\therefore B=\frac{-2}{5}$

$Put x=1$

$3(1)-1=C(1-4)(1+1)$ $\therefore 2=C(-3)(2)$

$\therefore C=\frac{-1}{3}$

$\therefore \frac{3 x-1}{(x-4)(x+1)(x-1)}=\frac{\frac{11}{15}}{x-4}+\frac{\frac{-2}{5}}{x+1}+\frac{\frac{-1}{3}}{x-1}$