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Integrate the function
$\frac{1}{x(\log x)^m}, x>0$
$\frac{1}{x(\log x)^m}, x>0$
Solution:
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Verified Answer
Put $\log \mathrm{x}=\mathrm{t}$, so that $\frac{1}{x} d x=d t$
$\begin{aligned}
&\therefore \quad \int \frac{1}{x(\log x)^m} d x=\int \frac{d t}{t^m}=\frac{t^{-m+1}}{-m+1}+\mathrm{C} \\
&=\frac{(\log x)^{1-m}}{1-m}+\mathrm{C}
\end{aligned}$
$\begin{aligned}
&\therefore \quad \int \frac{1}{x(\log x)^m} d x=\int \frac{d t}{t^m}=\frac{t^{-m+1}}{-m+1}+\mathrm{C} \\
&=\frac{(\log x)^{1-m}}{1-m}+\mathrm{C}
\end{aligned}$
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