Let $I=\int \frac{x}{(x^2+1)(x-1)}dx$ Let $\frac{x}{(x^2+1)(x-1)}$ $=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$ $\Rightarrow x = A (x^2 + 1) + (Bx + C) (x-1)$ ...(i) Putting x = 1 in (i), we get 1 = A(1 + 1) ⇒ A $=\frac{1}{2}$ Comparing coefficient of $x^2$ in (i), we get, 0 = A + B ⇒ B $=-\frac{1}{2}$ Putting x = 0 in (i), we get, 0 = A – C ⇒ C $=\frac{1}{2}$$\therefore \int \frac{x}{(x^2+1)(x-1)}dx$ $=\int \left[ \frac{\frac{1}{2}}{x-1}+\frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1} \right]dx$$=\frac{1}{2}\int \frac{dx}{x-1}-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{dx}{x^2+1}$$=\frac{1}{2}\int \frac{dx}{x-1}-\frac{1}{4}\int \frac{2x}{x^2+1}dx+\frac{1}{2}\int \frac{dx}{x^2+1}$