Answer
The required solution is $\frac{1}{\left( x+h+1 \right)\left( x+1 \right)}$.
Work Step by Step
We have the given complex rational expression:
$\frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}$.
We know that a rational expression is one that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator, $q\ne 0$.
Furthermore, a complex rational expression or a complex fraction is an algebraic rational expression in which either the numerator or denominator or both the numerator and denominator contain rational expressions.
And the given complex rational expression is
$\frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}$.
The numerator is $\frac{x+h}{x+h+1}-\frac{x}{x+1}$. Simplify the numerator:
$\begin{align}
& \frac{x+h}{x+h+1}-\frac{x}{x+1}=\frac{x+h}{x+h+1}\times \frac{\left( x+1 \right)}{\left( x+1 \right)}-\frac{x}{x+1}\times \frac{\left( x+h+1 \right)}{\left( x+h+1 \right)} \\
& =\frac{\left( x+h \right)\left( x+1 \right)}{\left( x+h+1 \right)\left( x+1 \right)}-\frac{x\left( x+h+1 \right)}{\left( x+1 \right)\left( x+h+1 \right)} \\
& =\frac{\left( x+h \right)\left( x+1 \right)-x\left( x+h+1 \right)}{\left( x+h+1 \right)\left( x+1 \right)} \\
& =\frac{{{x}^{2}}+x+xh+h-{{x}^{2}}-xh-x}{\left( x+h+1 \right)\left( x+1 \right)}.
\end{align}$.
Simplifying further,
$\frac{x+h}{x+h+1}-\frac{x}{x+1}=\frac{h}{\left( x+h+1 \right)\left( x+1 \right)}$.
And simplify the given complex rational expression:
$\begin{align}
& \frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}=\frac{\frac{h}{\left( x+h+1 \right)\left( x+1 \right)}}{h} \\
& =\frac{1}{\left( x+h+1 \right)\left( x+1 \right)}
\end{align}$.
Thus, $\frac{\frac{x+h}{x+h+1}-\frac{x}{x+1}}{h}=$ $\frac{1}{\left( x+h+1 \right)\left( x+1 \right)}$ ; $x\ne -1\ ;\ h\ne -\left( x+1 \right)$.
