Answer
$\displaystyle \frac{1}{3}\ln|\ln x|+C$
Work Step by Step
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
$=\displaystyle \frac{1}{3}\ln|u|+C$
$=\displaystyle \frac{1}{3}\ln|\ln x|+C$
