Definite integral via u-substitution with u = ln x

Let $u = \ln x$, so $du = \frac{1}{x}dx$. The integral $\int \frac{1}{x \ln x}dx$ becomes $\int \frac{1}{u} du$. Now update the bounds: when $x = e$, $u = \ln e = 1$; when $x = e^2$, $u = \ln(e^2) = 2$.

The integral transforms into $\int_{1}^{2} \frac{1}{u} du = \left[\ln|u|\right]_1^2 = \ln 2 - \ln 1 = \ln 2 - 0 = \ln 2$. So the value of the definite integral is $\ln 2$.