Answer
$\frac{1}{(x+1)(x+h+1)}; x \ne -1, x\ne-h-1, h \ne 0;$
Work Step by Step
$\frac{\frac{x+h}{x+h+1}- \frac{x}{x+1}}{h}$
Take LCD in the numerator,
$=\frac{\frac{(x+h)(x+1)-(x)(x+h+1)}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$
$=\frac{\frac{x^{2}+hx+x+h-(x^{2}+hx+x)}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$
$=\frac{\frac{x^{2}+hx+x+h-x^{2}-hx-x}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$
$=\frac{\frac{h}{(x+h+1)(x+1)}}{h}; x \ne -1, x\ne-h-1, h \ne 0;$
$=\frac{h}{(x+h+1)(x+1)} \times \frac{1}{h} ; x \ne -1, x\ne-h-1, h \ne 0;$
$= \frac{1}{(x+h+1)(x+1)}; x \ne -1, x\ne-h-1, h \ne 0;$
