Answer
$\frac{4x+16}{(x+3)^{2}}; x \ne -3;$ or $\frac{4(x+4)}{(x+3)^{2}}; x \ne -3;$
Work Step by Step
$\frac{4}{x^{2}+6x+9} + \frac{4}{x+3}$
$x^{2}+6x+9$ is in the form of $A^{2}+2AB+B^{2} = (A+B)^{2}$
$x^{2}+6x+9 = (x+3)^{2}$ So,
$=\frac{4}{ (x+3)^{2}} + \frac{4}{x+3}$
Taking Least Common Denominator,
$=\frac{4+4(x+3)}{ (x+3)^{2}} ; x \ne -3 $
$=\frac{4+4x+12}{ (x+3)^{2}} ; x \ne -3 $
$= \frac{4x+16}{(x+3)^{2}}; x \ne -3;$
$=\frac{4(x+4)}{(x+3)^{2}}; x \ne -3;$
