$\int\left(\frac{1}{\log_ex}-\frac{1}{(\log_ex)^2}\right)dx$ is equal to

$\frac{x}{\log_ex}+c$: C is an arbitrary constant

The correct answer is Option (1) → $\frac{x}{\log_ex}+c$: C is an arbitrary constant

Given integral:

$\int\left(\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2}\right)dx$

Let $\log_e x = t \Rightarrow \frac{dx}{x}=dt \Rightarrow dx = e^t dt$.

Substituting $x=e^t$:

$\int\left(\frac{1}{t}-\frac{1}{t^2}\right)e^t dt$

$\frac{d}{dt}\left(\frac{e^t}{t}\right)=\frac{e^t(t-1)}{t^2}=\frac{e^t}{t}-\frac{e^t}{t^2}$

Thus, $\int\left(\frac{e^t}{t}-\frac{e^t}{t^2}\right)dt=\frac{e^t}{t}+C$

Substitute back $t=\log_e x$ and $e^t=x$:

$\int\left(\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2}\right)dx=\frac{x}{\log_e x}+C$