Question:
Evaluate: $\int \frac{1}{\mathrm{x}(1+\log \mathrm{x})} \mathrm{dx}$
Solution:
Given, $\int \frac{1}{x(1+\log x)} d x$
Let $1+\log x=t$
$\Rightarrow \frac{d}{d x}(1+\log x)=d t$
$\Rightarrow \frac{1}{x} d x=d t$
$=\int \frac{1}{t} d t$
$=\log t+c$
$=\log (1+\log x)+c$