Answer:
Let \[l=\int{\frac{1}{x\,{{(\log x)}^{n}}}dx}\] Put \[t=\log x\] \[\Rightarrow \] \[dt=\frac{1}{x}dx\] \[\therefore \] \[l=\int{\frac{1}{{{(t)}^{n}}}dt=\frac{{{t}^{-\,n+1}}}{-\,n+1}}+C\] \[=\frac{{{(\log x)}^{1\,\,-\,\,n}}}{1-n}+C\] \[[\because t=\log x]\]
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