The value of $\int\left\{\frac{1}{\log_ex}-\frac{1}{(\log_ex)^2}\right\} dx$ is

$\frac{x}{\log_e x} + c$: where c is an arbitrary constant

The correct answer is Option (4) → $\frac{x}{\log_e x} + c$: where c is an arbitrary constant

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

Put $t=\log_e x$ so that $dx=e^t dt$ and $x=e^t$.

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

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

Derivative of $\frac{e^t}{t}$:

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

Thus the integrand equals $\frac{d}{dt}\left(\frac{e^t}{t}\right)$.

$\text{ therefore }\;\int\left(\frac{1}{t}-\frac{1}{t^2}\right)e^t dt=\frac{e^t}{t}+C$

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

$=\frac{x}{\log_e x}+C$

The value of the integral is $\frac{x}{\log_e x}+C$.