Question

$\int \frac{1}{x(3+\log x)}dx$

Answer 1

$$\begin{gathered} \mathrm{Here},\mathrm{we}\mathrm{are}\mathrm{considering}\log x\mathrm{as}{\log }_{\mathrm{e}}x. \\ \mathrm{Let}I=\int \frac{1}{x\left(3+\log x\right)}dx \\ \mathrm{Putting}\log x=t \\ \Rightarrow \frac{1}{x}=\frac{dt}{dx} \\ \Rightarrow \frac{dx}{x}=dt \\ \therefore I=\int \frac{dt}{3+t} \\ =\log \left|3+t\right|+C \\ =\log \left|3+\log x\right|+C\left[\because t=\log x\right] \end{gathered}$$