Evaluate $\int \left\{ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right\} dx$ (where $x > 1$).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{x}{\log x} + C$
$\int \left\{ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right\} dx$
$= \int \frac{1}{\log x} dx - \int \frac{1}{(\log x)^2} dx$
$= \frac{1}{\log x} \cdot \int 1 dx - \int \left\{ \frac{d}{dx} \left( \frac{1}{\log x} \right) \int 1 dx \right\} dx - \int \frac{1}{(\log x)^2} dx$
$= \frac{x}{\log x} + \int \frac{1}{(\log x)^2} \cdot \frac{1}{x} \cdot x dx - \int \frac{1}{(\log x)^2} dx$
$= \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx - \int \frac{1}{(\log x)^2} dx$
$= \frac{x}{\log x} + C$