Question

Write a value of $\int \frac{1+\log x}{3+x\log x}dx$.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{1+\log x}{3+x\log x}\right)dx \\ \mathrm{Let}3+x\log x=t \\ \Rightarrow 0+\left(x.\frac{1}{x}+\log x\right)dx=dt \\ \Rightarrow \left(1+\log x\right)dx=dt \\ \therefore I=\int \frac{dt}{t} \\ =\log t+C \\ =\log \left(3+x\log x\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }t\mathit{=}3+x\mathit{}\log \mathit{}x\right) \end{gathered}$$