Answer
$\frac{1}{x+3}$.
Work Step by Step
The given expression is
$=\frac{\frac{1}{x+2}}{1+\frac{1}{x+2}}$
Multiply the numerator and the denominator by $(x+2)$.
$=\frac{(x+2)}{(x+2)}\cdot \frac{\frac{1}{x+2}}{1+\frac{1}{x+2}}$
Use the distributive property.
$=\frac{(x+2)\cdot\frac{1}{x+2}}{(x+2)\cdot (1)+(x+2)\cdot \frac{1}{x+2}}$
Simplify.
$=\frac{1}{x+2+1}$
$=\frac{1}{x+3}$.
