Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2}\right)dx \\ \mathrm{Put}\log x=t \\ \Rightarrow x=e^t \\ \Rightarrow dx=e^{t}dt \\ \therefore I=\int e^t\left(\frac{1}{t}-\frac{1}{t^2}\right)dt \\ \mathrm{Here},f\left(t\right)=\frac{1}{t} \\ \Rightarrow f\text{'}\left(t\right)=\frac{-1}{t^2} \\ \mathrm{let}{e}^t\times \frac{1}{t}=p \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}t \\ \left(e^t\times \frac{1}{t}+e^t\times \frac{-1}{t^2}\right)dt=dp \\ \therefore I=\int dp \\ =p+C \\ =\frac{e^t}{t}+C \\ =\frac{x}{\log x}+C \end{gathered}$$