Question
Integrate the function: $\frac{1}{{x + x\log x}}$

Answer

Putting 1 + log x = t
$ \Rightarrow \frac{1}{x} = {dt}$
$ \Rightarrow \frac{{dx}}{x} = dt$
$\therefore \int {\frac{1}{{x + x\log x}}dx} $
$= \int {\frac{1}{{1 + \log x}}\frac{{dx}}{x}} $
$= \int {\frac{1}{t}dt = \log \left| t \right| + c} $
= log |1 + log x| + c

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