Definite integral via u-substitution

Let $u = \ln x$. Then $du = \frac{1}{x}dx$, so the integral becomes $\int u^2\, du$. Converting the bounds: when $x=1$, $u=\ln 1 = 0$; when $x=e$, $u=\ln e = 1$. This transforms the integral into $\int_0^1 u^2\,du$.

Now evaluate $\int_0^1 u^2\,du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$. So the value of the original definite integral is $\frac{1}{3}$.

When x equals e, since u equals the natural log of x, the new upper bound becomes the natural log of e, which equals one. This was shown when converting the bounds before integrating in terms of u. The lower bound similarly became zero, but the question asks specifically for the upper bound. So the answer is one.