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Question Number 107737 by mathdave last updated on 12/Aug/20

Answered by hgrocks last updated on 12/Aug/20

$$\ln (1-\mathrm{x})\ln (1+\mathrm{x})=\frac{1}{2}.((\ln (1+\mathrm{x})+$$$$\mathrm{l}{\mathrm{n}(1-\mathrm{x}))}^2-{\ln }^2(1+\mathrm{x})-{\ln }^2(1-\mathrm{x}))$$$$$$$$=\frac{1}{2}[{\ln }^2(1-{\mathrm{x}}^2)-{\ln }^2(1-\mathrm{x})-{\ln }^2(1+\mathrm{x})]$$$$$$$$[2\mathrm{ab}={(\mathrm{a}+\mathrm{b})}^2-{\mathrm{a}}^2-{\mathrm{b}}^2]$$$$[\mathrm{lnx}+\mathrm{lny}=\ln \left(\mathrm{xy}\right)]$$

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