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

Given integral: $\int \left(\frac{1}{\log_e t} - \frac{1}{(\log_e t)^2}\right) dt$

The integrand can be recognized as the derivative of $\frac{t}{\log_e t}$.

Consider differentiating $\frac{t}{\log_e t}$ using the quotient rule.

For $y = \frac{u}{v}$, the derivative is $\frac{dy}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2}$

Here, $u = t$ and $v = \log_e t$

So $\frac{du}{dt} = 1$ and $\frac{dv}{dt} = \frac{1}{t}$

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

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

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

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

This matches the given integrand exactly.

Therefore: $\int \left(\frac{1}{\log_e t} - \frac{1}{(\log_e t)^2}\right) dt = \frac{t}{\log_e t} + C$