Question:
$\frac{1}{x+x \log x}$
Solution:
$\frac{1}{x+x \log x}=\frac{1}{x(1+\log x)}$
Let $1+\log x=t$
$\therefore \frac{1}{x} d x=d t$
$\Rightarrow \int \frac{1}{x(1+\log x)} d x=\int_{t}^{1} d t$
$=\log |t|+\mathrm{C}$
$=\log |1+\log x|+\mathrm{C}$
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