The value of the integral $\int\limits_1^{e^6}\left[\frac{\log x}{3}\right] d x$, where [.] denotes the greatest integer function, is

$e^6-e^3$

We have,

$0<\frac{\log _e x}{3}<1$ when $1<x<e^3$

and, $1 \leq \frac{\log _e x}{3}<2$ when $e^3<x<e^6$

∴  $\left[\frac{\log _e x}{3}\right]= \begin{cases}0, & \text { when } 1<x<e^3 \\ 1, & \text { when } e^3<x<e^6\end{cases}$

$\Rightarrow \int\limits_1^{e^6}\left[\frac{\log _e x}{3}\right] d x=\int\limits_1^{e^3} 0 d x+\int\limits_{e^3}^{e^6} 1 d x=e^6-e^3$