Answer
$\frac{1}{(x+1)(x+h+1)}$
Work Step by Step
$\dfrac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}=\dfrac{1-\frac{1}{x+h+1}-(1-\frac{1}{x+1})}{h}=\dfrac{\frac{1}{x+1}-\frac{1}{x+h+1}}{h}=\dfrac{\frac{x+h+1}{(x+1)(x+h+1)}-\frac{x+1}{(x+1)(x+h+1)}}{h}=\dfrac{\frac{h}{(x+1)(x+h+1)}}{h}=\frac{1}{(x+1)(x+h+1)}$