Answer
$\frac{3x^2+8x+6}{(x+2)(x+3)^2}$.
Work Step by Step
The given expression is
$=\frac{2}{x^2+5x+6}+\frac{3x}{x^2+6x+9}$
Factor $x^2+5x+6$.
Rewrite the middle term $5x$ as $3x+2x$.
$\Rightarrow x^2+3x+2x+6$
Group the terms.
$\Rightarrow (x^2+3x)+(2x+6)$
Factor each group.
$\Rightarrow x(x+3)+2(x+3)$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x+2)$
Factor $x^2+6x+9$.
Rewrite the middle term $6x$ as $3x+3x$.
$\Rightarrow x^2+3x+3x+9$
Group the terms.
$\Rightarrow (x^2+3x)+(3x+9)$
Factor each group.
$\Rightarrow x(x+3)+3(x+3)$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x+3)$
Substitute back the factors into the given function.
$=\frac{2}{(x+3)(x+2)}+\frac{3x}{(x+3)(x+3)}$
The LCD is $(x+2)(x+3)(x+3)$.
$=\frac{2(x+3)}{(x+2)(x+3)(x+3)}+\frac{3x(x+2)}{(x+2)(x+3)(x+3)}$
Use the distributive property.
$=\frac{2x+6}{(x+2)(x+3)(x+3)}+\frac{3x^2+6x}{(x+2)(x+3)(x+3)}$
Add numerators because denominators are the same.
$=\frac{2x+6+3x^2+6x}{(x+2)(x+3)(x+3)}$
Simplify.
$=\frac{3x^2+8x+6}{(x+2)(x+3)(x+3)}$
$=\frac{3x^2+8x+6}{(x+2)(x+3)^2}$.
