Chapter 11 - Section 11.5 - Derivatives of Logarithmic and Exponential Functions - Exercises - Page 842: 26

$ \displaystyle \frac{1+3.1x^{-2}}{x-3.1x^{-1}}$

(see p. 838, Generalized Rule)$\quad$ $\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$ $u(x)=x-3.1x^{-1}$ $\displaystyle \frac{du}{dx}=1-3.1(-x^{-2})=1+3.1x^{-2}$ $\displaystyle \frac{d}{dx}[\ln|2x^{2}+1|]=\frac{1}{x-3.1x^{-1}}\cdot(1+3.1x^{-2})$ $=\displaystyle \frac{1+3.1x^{-2}}{x-3.1x^{-1}}$