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Question Number 172839 by DAVONG last updated on 02/Jul/22

$$\mathrm{I}=\int \frac{1}{\mathrm{lnx}}\mathrm{dx}=?$$

Answered by puissant last updated on 02/Jul/22

$$I=\int \frac{1}{lnx}dx;x=e^u\Rightarrow dx=e^{u}du$$$$I=\int \frac{e^u}{u}du=\int \sum _{n⩾0}\frac{u^{n-1}}{n!}du$$$$=\int (\frac{1}{u}+\sum _{n⩾1}\frac{u^{n-1}}{n!})du$$$$=lnu+\sum _{n⩾1}\frac{u^n}{n.n!}+C$$$$=ln\left(lnx\right)+\sum _{n⩾1}\frac{{ln}^{n}x}{n.n!}+C$$$$.......\mathcal{L}epuissant......$$

Commented by ghakhan last updated on 03/Jul/22

$$canyouexplainhowdidget\frac{1}{u}inthethirdstepplease?$$

Commented by puissant last updated on 03/Jul/22

$$\sum _{n⩾0}\frac{u^{n-1}}{n!}=\frac{u^{0-1}}{0!}+\sum _{n⩾1}\frac{u^{n-1}}{n!}$$$$=\frac{1}{u}+\sum _{n⩾1}\frac{u^{n-1}}{n!}.$$

Commented by DAVONG last updated on 03/Jul/22

$$\mathrm{Thanks}\mathrm{teacher}!$$

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