Solution :
We have, I = \(\int\) \(1\over x log x\) dx
Put log x = t \(\implies\) \(1\over x\) dx = dt
I = \(\int\) \(1\over t\) dt
I = log | t | + C
I = log |log x| + C
Hence, the integration of \(1\over x log x\) is log (log x) + C
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