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$\int \frac{\log x d x}{x^3}=$
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$-\frac{1}{4 x^2}(2 \log x+1)+c$
$\begin{aligned} & \int \frac{\log x}{x^3} d x=\int x^{-3} \log x d x \\ & =-\frac{\log x}{2 x^2}+\int \frac{1}{x} \cdot \frac{1}{2 x^2}+c=-\frac{\log x}{2 x^2}+\frac{1}{2} \cdot \frac{x^{-2}}{-2}+c \\ & =-\frac{\log x}{2 x^2}-\frac{1}{4 x^2}+c=-\frac{1}{4 x^2}(2 \log x+1)+c \\ & \end{aligned}$
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