Let $I=\int\frac{x}{(x-1)(x-2)(x-3)}dx$ 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) ...(i)Putting x = 1 in (i), we get 1 = A(1 – 2) (1 – 3) ⇒ A $=\frac{1}{2}$Putting x = 2 in (i), we get 2 = B(2 – 1) (2 – 3) ⇒ B = – 2Putting x = 3 in (i), we get 3 = C (3 – 1) (3 – 2) ⇒ C = $\frac{3}{2}$$\therefore\frac{x}{(x-1)(x-2)(x-3)}$ $=\frac{1}{2(x-1)}-\frac{2}{x-2}+\frac{3}{2(x-3)}$$\therefore\int\frac{x}{(x-1)(x-2)(x-3)}dx$ $=\frac{1}{2}\int\frac{1}{x-1}dx-2\int\frac{1}{x-2}+\frac{3}{2}\int\frac{1}{x-3}$