## Calculus 10th Edition

$\displaystyle \frac{1}{3}\ln|\ln x|+C$
First, $\ln x^{3}=3\ln x$ , so we rewrite the integral $...=\displaystyle \frac{1}{3}\int\frac{1}{\ln x}\cdot\frac{1}{x}dx, \quad\left[\begin{array}{l} u=\ln x\\ du=\frac{1}{x}dx \end{array}\right]$ $=\displaystyle \frac{1}{3}\int\frac{1}{u}du$= ... use Th.5.5.2 $=\displaystyle \frac{1}{3}\ln|u|+C$ $=\displaystyle \frac{1}{3}\ln|\ln x|+C$