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

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

Consider $y = \frac{x}{\log_e x}$ and differentiate using the quotient rule.

For $y = \frac{u}{v}$ where $u = x$ and $v = \log_e x$:

$\frac{du}{dx} = 1$

$\frac{dv}{dx} = \frac{1}{x}$

Applying the quotient rule:

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

$= \frac{\log_e x - 1}{(\log_e x)^2}$

Splitting the fraction:

$= \frac{\log_e x}{(\log_e x)^2} - \frac{1}{(\log_e x)^2}$

$= \frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}$

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

Therefore:

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

where $c$ is an arbitrary constant.