$\int \frac{1}{x (log x) log (log x)} d x =$

log (log x) + C

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log | log (x log x) + C

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log | log | log (log x) | | + C

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log | x log (log x) | + C

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log | log (log x) | + C

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Solution

The correct option is E $log | log (log x) | + C$
Let $l = \int \frac{d x}{(log x) \cdot x log (log x)}$
Put, $t = log (log x)$
$\Rightarrow \frac{d t}{d x} = \frac{1}{x log x}$
$\Rightarrow x log x d t = d x$
Then, $l = \int \frac{d t}{t} = log | t | + C$
$= log | log (log x) | + C$

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