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Question Number 130907 by greg_ed last updated on 30/Jan/21

$${\int }_0^1\frac{\ln (1+x+x^2)}{x}dx$$

Commented by benjo_mathlover last updated on 30/Jan/21

այս հարցն արդեն պատասխանվել է

Commented by benjo_mathlover last updated on 30/Jan/21

$$\mathrm{qn}130676$$

Answered by Dwaipayan Shikari last updated on 30/Jan/21

$${\int }_0^1\frac{log(1-x^3)}{x}-\frac{log(1-x)}{x}dx$$$$=-\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{x^{3n-1}}{n}+\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{x^{n-1}}{n}dx$$$$=-\frac{1}{3}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\zeta \left(2\right)(1-\frac{1}{3})=\frac{{\pi }^2}{9}$$

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