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Question Number 64477 by Chi Mes Try last updated on 18/Jul/19

$$plsineediturgently...amstuck$$$$workingsplease$$$$\left(1\right)\int Ln(1-Lnx)dx$$$$$$$$\left(2\right)\int \frac{1}{Lnx}dx$$$$$$$$\left(3\right)\int Ln(-2Lnx)dx$$$$$$$$Godwillhonouru4urreplies$$

Commented by Prithwish sen last updated on 18/Jul/19

$$1)\int \frac{\mathrm{xLn}(1-\mathrm{Lnx})\mathrm{dx}}{\mathrm{x}}\mathrm{put}1-\mathrm{Lnx}=\mathrm{t}$$$$\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{dt}\mathrm{and}\mathrm{x}={\mathrm{e}}^{1-\mathrm{t}}\Rightarrow -\int {\mathrm{e}}^{1-\mathrm{t}}\mathrm{Lnt}\mathrm{dt}$$$$\mathrm{now}\mathrm{take}\mathrm{int}.\mathrm{by}\mathrm{parts}$$$$2)\int \frac{\mathrm{xdx}}{\mathrm{xLnx}}\mathrm{putLnx}=\mathrm{t}\mathrm{and}\mathrm{you}\mathrm{will}\mathrm{get}$$$$\int \frac{{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}}{\mathrm{t}}\mathrm{and}\mathrm{then}\mathrm{by}\mathrm{parts}$$$$3)\int \mathrm{Ln}(-2\mathrm{Lnx})\mathrm{dx}\mathrm{put}\mathrm{Lnx}=-\frac{\mathrm{t}}{2}$$$$\mathrm{and}\mathrm{you}\mathrm{get}-\frac{1}{2}\int {\mathrm{e}}^{\frac{-\mathrm{t}}{2}}\mathrm{Lnt}\mathrm{dt}\mathrm{and}\mathrm{again}\mathrm{take}$$$$\mathrm{by}\mathrm{parts}.$$

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