Answer
$\displaystyle \frac{4(x+4)}{(x+3)^{2}},\quad x\neq-3$
Work Step by Step
Factor each denominator.
Recognize a perfect square in the first.
$\displaystyle \frac{4}{x^{2}+6x+9}+\frac{4}{x+3}=\frac{4}{(x+3)^{2}}+\frac{4}{x+3}=...$
The denominators are different. Find a common denominator.
LCD=$(x+3)^{2}$
$=\displaystyle \frac{4}{(x+3)^{2}}+\frac{4}{x+3}\times\frac{x+3}{x+3}$
$=\displaystyle \frac{4}{(x+3)^{2}}+\frac{4x+12}{(x+3)^{2}}$
$=\displaystyle \frac{4+4x+12}{(x+3)^{2}}$
$=\displaystyle \frac{4x+16}{(x+3)^{2}}$
= $\displaystyle \frac{4(x+4)}{(x+3)^{2}},\quad x\neq-3$
