$\int \dfrac{x}{(x-1)(x-2)}$ $dx$

The correct answer is $log \dfrac{(x-2)^2}{|x-1|} + C$

Let $ \dfrac{x}{(x-1)(x-2)}$ = $\dfrac{A}{x - 1}$ + $\dfrac{B}{x - 2}$

⇒ $x$ = $A(x-2) + B(x-1)$

Putting $x$ = 1 in above equation, we get $A$ = -1

Putting $x$ = 2 in above equation, we get $B$ = 2


∴ $\int \dfrac{x}{(x-1)(x-2)}$ $dx$ = $\int \left( \dfrac{-1}{x - 1} + \dfrac{2}{x - 2} \right)$ $dx$

= $-log|x-1| + 2\log|x - 2|+C$

= $log \dfrac{(x-2)^2}{|x-1|} + C$