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Question Number 152160 by peter frank last updated on 26/Aug/21

$$\int [\frac{1}{\log \mathrm{x}}-\frac{1}{{\left(\log \mathrm{x}\right)}^2}]\mathrm{dx}$$

Answered by Olaf_Thorendsen last updated on 26/Aug/21

$$\mathrm{F}\left(x\right)=\int (\frac{1}{\log x}-\frac{1}{{\log }^{2}x})dx$$$$\mathrm{F}\left(x\right)=\int \frac{\log x-1}{{\log }^{2}x}dx$$$$\mathrm{F}\left(x\right)=\int \frac{d}{dx}\left(\frac{x}{\log x}\right)dx$$$$\mathrm{F}\left(x\right)=\frac{x}{\log x}+\mathrm{C}$$

Commented by puissant last updated on 26/Aug/21

$$jolie!$$

Commented by peter frank last updated on 26/Aug/21

$$\mathrm{thank}\mathrm{you}$$

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