Answer
$ \displaystyle \frac{2x-0.63x^{-0.7}}{x^{2}-2.1x^{0.3}}$
Work Step by Step
(see p. 838, Generalized Rule)$\quad$
$\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$
$u(x)=x^{2}-2.1x^{0.3}$
$\displaystyle \frac{du}{dx}=2x-2.1(0.3x^{-0.7})=2x-0.63x^{-0.7}$
$\displaystyle \frac{d}{dx}[\ln|2x^{2}+1|]=\frac{1}{x^{2}-2.1x^{0.3}}\cdot(2x-0.63x^{-0.7})$
$=\displaystyle \frac{2x-0.63x^{-0.7}}{x^{2}-2.1x^{0.3}}$
