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

$\frac{t}{\log_et}+C$, where C is constant of integration

The correct answer is Option (2) → $\frac{t}{\log_et}+C$, where C is constant of integration

Given:

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

Let $u=\log t$ so that $\frac{dt}{t}=du$ and $dt=t\,du=e^{u}du$.

Then the integral becomes:

$\displaystyle \int\left(\frac{1}{u}-\frac{1}{u^{2}}\right)e^{u}du$

Recognize derivative:

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

Hence:

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

Substitute back $u=\log t$:

$=\frac{t}{\log t}+C$

Final answer: $\frac{t}{\log t}+C$