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integrates 1/x(logx)powerof n w.r.t.x
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u=ln(x)
\[\int\limits \frac{1}{ x [\ln(x)]^n} dx\]
if its that^
if its that^
let u=ln(x) du = (1/x) dx
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\[= \int\limits \frac{1}{u^n} du \]
\[= \int\limits u^{-n} du = \frac{u^{-n+1} }{1-n} +C \]
but that is only for \[n \neq \]
\[= \frac{ (\ln(x))^{1-n} }{ 1-n} +C \] for \[n \neq 1\] = \[\ln(\ln(x)) +C \] for n=1
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