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

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


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

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

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