Chapter 5 - Exponential and Logarithmic Functions - 5.2 One-to-One Functions; Inverse Functions - 5.2 Assess Your Understanding - Page 265: 4

$\frac{x}{1-x}$

$\dfrac{\frac{1}{x}+1}{\frac{1}{x^2}-1}=\dfrac{\frac{1}{x}+\frac{x}{x}}{\frac{1}{x^2}-\frac{x^2}{x^2}}=\dfrac{\frac{x+1}{x}}{\frac{1-x^2}{x^2}}=\frac{x+1}{x}\cdot\frac{x^2}{1-x^2}=(x+1)\cdot\frac{x}{(1+x)(1-x)}=\frac{x}{1-x}$