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

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

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

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

⇒ $x$ = $A (x+2) + B(x+1)$


For $x= -1$,

$-1$ = $A(-1 + 2) + B(-1+1)$

⇒ $A=-1$

For $x= -2$,

$-2$ = $A(-2 + 2) + B(-2+1)$

⇒ $B=2$



Putting values of $A$ and $B$ in equation (1) and integrating,

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

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

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

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