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$\int [\log (\log x) + \frac{1}{{(\log x)}^{2}}] d x =$

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loglogx+xlogx+c

xloglogx+1logx+c

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−loglogx−xlogx+c

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detailed solution

Correct option is C

∫loglogx+1logx2dx=∫logt+1t2etdt put logx=t then x=et ⇒dx=etdt∫logt-1t+1t+1t2etdt =logt-1tet =xloglogx-1logx+c

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