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Question Number 147310 by mnjuly1970 last updated on 19/Jul/21

Commented by Tawa11 last updated on 03/Aug/21

$$\mathrm{Great}$$

Answered by qaz last updated on 20/Jul/21

$${\int }_0^1\ln \left(\mathrm{x}\right)\cdot \ln (1+\mathrm{x})\mathrm{dx}$$$$=\mathrm{xln}\left(\mathrm{x}\right)\cdot \ln (1+\mathrm{x}){\mid }_0^1-{\int }_0^1\mathrm{x}(\frac{\ln (1+\mathrm{x})}{\mathrm{x}}+\frac{\mathrm{lnx}}{1+\mathrm{x}})\mathrm{dx}$$$$=-{\int }_0^1(\ln (1+\mathrm{x})+(1-\frac{1}{1+\mathrm{x}})\mathrm{lnxdx}$$$$=-{[(1+\mathrm{x})\ln (1+\mathrm{x})-(1+\mathrm{x})]}_0^1-{\int }_0^1\mathrm{lnxdx}+{\int }_0^1\frac{\mathrm{lnx}}{1+\mathrm{x}}\mathrm{dx}$$$$=1-2\ln 2-{[\mathrm{xlnx}-\mathrm{x}]}_0^1+\ln (1+\mathrm{x})\mathrm{lnx}{\mid }_0^1-{\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=2-2\ln 2-\frac{{\pi }^2}{12}$$

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