$\int \dfrac{2x}{x^2 + 3x + 2}$ $dx$

The correct answer is $4log|x + 2| - 2log|x + 1|+C$

Let $ \dfrac{2x}{x^2 + 3x + 2}$ = $\dfrac{2x}{(x + 2)(x + 1)}$ = $\dfrac{A}{x + 2}$ + $\dfrac{B}{x + 1}$

⇒ $2x$ = $A(x+1) + B(x+2)$

Putting $x$ = -2 in above equation, we get $A$ = 4

Putting $x$ = -1 in above equation, we get $B$ = -2


∴ $ \int \dfrac{2x}{x^2 + 3x + 2}$ $dx$= $\int \left( \dfrac{4}{x + 2} + \dfrac{-2}{x + 1} \right)$ $dx$

= $4log|x+2| - 2log|x + 1|+C$