Question

The value of $\int \frac{1}{x+x\log x}dx$ is
(a) 1 + log x
(b) x + log x
(c) x log (1 + log x)
(d) log (1 + log x)

Answer 1

(d) log (1 + log x)

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{x+x\log x} \\ \Rightarrow \int \frac{dx}{x\left(1+\log x\right)} \\ \mathrm{Putting}1+\log x=t \\ \Rightarrow \frac{1}{x}dx=dt \\ \therefore I=\int \frac{dt}{t} \\ =\ln \left|t\right|+C \\ =\ln \left|1+\log x\right|+C \end{gathered}$$