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

The integral to evaluate is $\int \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right)dx$.

Since the integrand contains $\log_e x$ terms in the denominator, consider whether $\frac{x}{\log_e x}$ could be the antiderivative.

Taking the derivative of $y = \frac{x}{\log_e x}$ using the quotient rule where $u = x$ and $v = \log_e x$:

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

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

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

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

Splitting the fraction:

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

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

This matches the integrand exactly.

Therefore:

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