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A question on my book: $$\int{\frac{1}{x(\log{x})^n}dx}$$where n is an integer.

What I did:
Let u=$\frac{1}{\log{x}}$, then $\frac{du}{dx}=x, dx=\frac{du}{x}.$

Now the integral becomes $$\int{\frac{1}{xu^n}dx}=\int{\frac{1}{xu^n}\times {\frac{du}{x}}}$$

I don't know what to do next. Should I change another u? Note: if possible, please use substitution to solve the problem.

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    $\begingroup$ $\dfrac{\mathrm d u}{\mathrm d x}=-\dfrac1{x\log^2x}$, not $\;\dfrac1{\log' x}$. $\endgroup$ – Bernard Oct 11 '17 at 22:10
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Consider the substitution $u=\log x$ and so $du=\frac{1}{x}dx$. Hence,$$ \int \frac{1}{x(\log x)^n}\,dx=\int\frac{(\log x)^{-n}}{x}\,dx=\int u^{-n}\,du. $$ Then, for $n\geq 2$, we get$$ \int u^{-n}\,du=\frac{u^{1-n}}{1-n}=\frac{(\log x)^{1-n}}{1-n}. $$ For $n=1$, we get$$ \int u^{-1}\,du=\log u=\log(\log x). $$

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    $\begingroup$ Not true if $n=1$. $\endgroup$ – Mark Viola Oct 11 '17 at 23:02
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Call $u=\log x$, then $du = dx/x$ and your integral becomes

$$ \int\frac{dx}{x\log^n x} = \int\frac{du}{u^n} = \cdots $$

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