Value of $\int\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right)dx$ is

$\frac{x}{\log x} + c$, where c is an arbitrary constant

The correct answer is Option (1) → $\frac{x}{\log x} + c$, where c is an arbitrary constant

$I = \int \left(\frac{1}{\log x} - \frac{1}{(\log x)^2}\right) \, dx$

Let $\log x = t \Rightarrow dx = e^t \, dt$.

$I = \int \frac{1}{t} e^t \, dt - \int \frac{1}{t^2} e^t \, dt$

Using integration by parts for the second term:

$\int \frac{1}{t^2} e^t \, dt = -\frac{e^t}{t} + \int \frac{e^t}{t} \, dt$

Combining both integrals:

$I = \int \frac{e^t}{t} \, dt - \left(-\frac{e^t}{t} + \int \frac{e^t}{t} \, dt\right)$

$= \frac{e^t}{t} = \frac{x}{\log x} + C$