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Question Number 166082 by mnjuly1970 last updated on 12/Feb/22

$$$$$$prove$$$$\mathrm{\Omega }={\int }_0^1\frac{{(1-x)}^2.{ln}^3(1-x)}{x}dx=\frac{51}{8}-\frac{{\pi }^4}{15}◼m.n$$$$$$$$$$

Answered by qaz last updated on 13/Feb/22

$${\int }_0^1\frac{{(\mathrm{l}-\mathrm{x})}^2{\ln }^3(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$={\int }_0^1\frac{{\mathrm{x}}^2{\ln }^3\mathrm{x}}{1-\mathrm{x}}\mathrm{dx}$$$$=-{\int }_0^1(1+\mathrm{x}){\ln }^3\mathrm{xdx}+{\int }_0^1\frac{{\ln }^3\mathrm{x}}{1-\mathrm{x}}\mathrm{dx}$$$$=-(\mathrm{x}+\frac{1}{2}{\mathrm{x}}^2){\ln }^3\mathrm{x}{\mid }_0^1+3{\int }_0^1(1+\frac{1}{2}\mathrm{x}){\ln }^2\mathrm{xdx}-\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{6}{{(\mathrm{n}+1)}^4}$$$$=3(\mathrm{x}+\frac{1}{4}{\mathrm{x}}^2){\ln }^2\mathrm{x}{\mid }_0^1-6{\int }_0^1(1+\frac{1}{4}\mathrm{x})\mathrm{lnxdx}-\frac{{\pi }^4}{15}$$$$=-6(\mathrm{x}+\frac{1}{8}{\mathrm{x}}^2)\mathrm{lnx}{\mid }_0^1+6{\int }_0^1(1+\frac{1}{8}\mathrm{x})\mathrm{dx}-\frac{{\pi }^4}{15}$$$$=6(\mathrm{x}+\frac{1}{16}{\mathrm{x}}^2){\mid }_0^1-\frac{{\pi }^4}{15}$$$$=\frac{51}{8}-\frac{{\pi }^4}{15}$$

Commented by mnjuly1970 last updated on 13/Feb/22

$$thxalotsirqaz$$

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