Here's a kick off for you

\(\displaystyle \int_{e}^{e^2} \frac {1-\ln{x}}{x\ln{x}} ~dx\)

\(\displaystyle u = \ln{x} \implies \frac{du}{dx}= \frac{1}{x}\)

Now \(\displaystyle \int\frac {1-\ln{x}}{x\ln{x}} ~dx = \int\frac{1}{x}\left(\frac {1-\ln{x}}{\ln{x}}\right) ~dx =\int \frac{1-u}{u}~du\)