Solution :

$\text{Let}$

$\text{By using partial fraction method ,}$

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

$ \begin{array}{I} \therefore 3 x-1=A(x+1)(x-1)+B(x-4)(x-1)+C(x-4)(x+1) \end{array} $

$ \begin{array}{I} \text{Putting x =4 to get A value, } \\ \therefore 3(4)-1 =A(4+1)(4-1) \\ \therefore 11 =15 A \\ \therefore A =\frac{11}{15} \\ \text{Putting x =-1 to get B value, } \\ \therefore3(-1)-1 =B(-1-4)(-1-1)\\ \therefore-4 =B(-5)(-2)\\ \therefore B =\frac{-2}{5}\\ \text{Putting x =1 to get C value, } \\ \therefore 3(1)-1 =C(1-4)(1+1)\\ \therefore 2 =C(-3)(2)\\ \therefore C =\frac{-1}{3}\\ \end{array} $

$ \begin {array}{I} \text{Hence , Resolved partial fraction is}\\ \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} \end{array} $