Integrate the rational function: $\frac{2x}{x^{2}+3x+2}$

Let $\frac{2x}{x^{2}+3x+2} = \frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.
Multiplying both sides by $(x+1)(x+2)$,we get:
$2x = A(x+2) + B(x+1)$ ........$(1)$
To find $A$,put $x = -1$ in equation $(1)$:
$2(-1) = A(-1+2) + B(0)$
$-2 = A(1) \implies A = -2$.
To find $B$,put $x = -2$ in equation $(1)$:
$2(-2) = A(0) + B(-2+1)$
$-4 = B(-1) \implies B = 4$.
Therefore,$\frac{2x}{(x+1)(x+2)} = \frac{-2}{x+1} + \frac{4}{x+2}$.
Integrating both sides with respect to $x$:
$\int \frac{2x}{x^{2}+3x+2} dx = \int \left( \frac{4}{x+2} - \frac{2}{x+1} \right) dx$
$= 4 \int \frac{1}{x+2} dx - 2 \int \frac{1}{x+1} dx$
$= 4 \log |x+2| - 2 \log |x+1| + C$,where $C$ is an arbitrary constant.