Evaluate the integral $I = \int_{2}^{e} \left( \frac{1}{\ln x} - \frac{1}{\ln^2 x} \right) dx$.
Step-by-Step Solution
Key Concept: General
Let $I = \int_{2}^{e} \left( \frac{1}{\ln x} - \frac{1}{\ln^2 x} \right) dx$<br>Put $\ln x = t \Rightarrow x = e^t \Rightarrow dx = e^t dt$<br>When $x = 2 \Rightarrow t = \ln 2$; when $x = e \Rightarrow t = 1$<br>$\therefore I = \int_{\ln 2}^{1} e^t \left( \frac{1}{t} + \frac{-1}{t^2} \right) dt = \left[ \frac{e^t}{t} \right]_{\ln 2}^{1} = e - \frac{2}{\ln 2}$
Correct Answer: $e - \frac{2}{\ln 2}$