## College Algebra 7th Edition

$\frac{1}{2x+1}$
We multiply by $x(x+1)$ to simplify the fractions: $\displaystyle \frac{\frac{1}{x}-\frac{1}{x+1}}{\frac{1}{x}+\frac{1}{x+1}}=\frac{\frac{1}{x}-\frac{1}{x+1}}{\frac{1}{x}+\frac{1}{x+1}}$ * $\displaystyle \frac{x(x+1)}{x(x+1)}=\frac{(x+1)-x}{(x+1)+x}=\frac{1}{2x+1}$