Answer
$\displaystyle \frac{4x}{2x^{2}+1}$
Work Step by Step
By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$
Here, u(x)$=2x^{2}+1,\displaystyle \ \ \ \frac{du}{dx}=4x$
$\displaystyle \frac{d}{dx}[\ln(2x^{2}+1)]=\frac{1}{2x^{2}+1}\cdot 4x=\frac{4x}{2x^{2}+1}$
