Let $I=\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx$ Let $\frac{3x-1}{(x-1)(x-2)(x-3)}$ $=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ Putting $x=1$ in (i), we get3 - 1 = A(1 - 2) (1- 3) ⇒ 2 = A (-1) (-2) ⇒ A = 1Putting x = 2 in (i), we get6 – 1 = B(2 – 1) (2 – 3) ⇒ 5 = B(1) (–1) ⇒ B = –5Putting x = 3 in (i), we get9 – 1 = C(3 – 1) (3 – 2) ⇒ 8 = C(2) (1) ⇒ C = 4$\therefore \frac{3x-1}{(x-1)(x-2)(x-3)}$ $=\frac{1}{x-1}-\frac{5}{x-2}+\frac{4}{x-3}$$\therefore \int \frac{3x-1}{(x-1)(x-2)(x-3)}$ $=\int \frac{1}{x-1}dx -5\int \frac{1}{x-2}dx +4\int \frac{1}{x-3}dx$