## Precalculus (6th Edition)

$\dfrac{x+1}{x-1}$
Simplify by multiplying the LCD (which is $x$) to both the numerator and the denominator to obtain: $=\dfrac{x(1+\frac{1}{x})}{x(1-\frac{1}{x})} \\=\dfrac{x(1)+x(\frac{1}{x})}{x(1)-x(\frac{1}{x})} \\=\dfrac{x+1}{x-1}$