Answer
$ \frac{x^2+1}{(x+1)x^3}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{\frac{1}{x+1}}{x-\frac{1}{x+\frac{1}{x}}}$
Solve the lowest denominator.
$=x+\frac{1}{x}$
$=\frac{x}{1}+\frac{1}{x}$
The LCD of the denominators is $x$.
$=\frac{x(x)}{x}+\frac{1}{x}$
$=\frac{x^2+1}{x}$
Back substitute into the fraction.
$=\frac{1}{\frac{x^2+1}{x}}$
Invert the divisor and multiply.
$=\frac{x}{x^2+1}$
Back substitute into the fraction.
$\Rightarrow \frac{x}{x-\frac{x}{x^2+1}}$
Now solve the denominator.
$=x-\frac{x}{x^2+1}$
$=\frac{x}{1}-\frac{x}{x^2+1}$
The LCD of the denominators is $x^2+1$.
$=\frac{x(x^2+1)}{x^2+1}-\frac{x}{x^2+1}$
$=\frac{x(x^2+1)-x}{x^2+1}$
Simplify.
$=\frac{x^3+x-x}{x^2+1}$
$=\frac{x^3}{x^2+1}$
Back substitute into the fraction.
$\Rightarrow \frac{\frac{1}{x+1}}{\frac{x^3}{x^2+1}}$
Invert the divisor and multiply.
$\Rightarrow \frac{1}{x+1}\cdot \frac{x^2+1}{x^3}$.
$\Rightarrow \frac{x^2+1}{(x+1)x^3}$.
