Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 28

$$s=\int_{a}^{b}\sqrt{ \frac{f^2(x)+f'^2(x)}{f^2(x)} }$$

Since $$y=\ln [f(x)]\to y'=\frac{f'(x)}{f(x)} $$ Then \begin{align*} s&=\int_{a}^{b}\sqrt{1+y'^2}dx\\ &= \int_{a}^{b}\sqrt{1+\frac{f'^2(x)}{f^2(x)} }\\ &=\int_{a}^{b}\sqrt{ \frac{f^2(x)+f'^2(x)}{f^2(x)} }\\ \end{align*}