Answer
$\dfrac{2}{x^{2}+4x+4}+\dfrac{1}{x+2}=\dfrac{x+4}{(x+2)^{2}}$
Work Step by Step
$\dfrac{2}{x^{2}+4x+4}+\dfrac{1}{x+2}$
Factor the denominator of the first fraction, which is a perfect square trinomial:
$\dfrac{2}{x^{2}+4x+4}+\dfrac{1}{x+2}=\dfrac{2}{(x+2)^{2}}+\dfrac{1}{x+2}=...$
Evaluate the sum of the two rational expressions and then simplify:
$...=\dfrac{2+(x+2)}{(x+2)^{2}}=\dfrac{x+4}{(x+2)^{2}}$
