Answer
$\frac{x(4x+1)}{(x+1)(x+2)(2x-3)}$
Work Step by Step
Step 1. Factor the denominators $x^2+3x+2=(x+1)(x+2)$ and $2x^2-x-3=(x+1)(2x-3)$, thus the Least Common Denominator (LCD) is $(x+1)(x+2)(2x-3)$
Step 2. We have $\frac{x}{x^2+3x+2}+\frac{2x}{2x^2-x-3}=\frac{x(2x-3)}{(x+1)(x+2)(2x-3)}+\frac{2x(x+2)}{(x+1)(x+2)(2x-3)}=\frac{2x^2-3x+2x^2+4x}{(x+1)(x+2)(2x-3)}=\frac{4x^2+x}{(x+1)(x+2)(2x-3)}=\frac{x(4x+1)}{(x+1)(x+2)(2x-3)}$
