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Question Number 114541 by mnjuly1970 last updated on 19/Sep/20

Answered by abdomsup last updated on 19/Sep/20

$$A={\int }_0^1\frac{x^{2}ln\left(x\right)}{1-x^2}dx\Rightarrow$$$$A={\int }_0^1{x}^{2}ln\left(x\right)\sum _{n=0}^{\mathrm{\infty }}{x}^{2n}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^1{x}^{2n+2}ln\left(x\right)dx$$$$u_n={\int }_0^1{x}^{2n+2}ln\left(x\right)dx$$$$={\left[\frac{x^{2n+3}}{2n+3}ln\left(x\right)\right]}_0^1-{\int }_0^1\frac{x^{2n+2}}{2n+3}dx$$$$=-\frac{1}{{(2n+3)}^2}{\left[x^{2n+3}\right]}_0^1=-\frac{1}{{(2n+3)}^2}$$$$\Rightarrow A=-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+3)}^2}$$$${=}_{n=p-1}-\sum _{p=1}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}$$$$wehave\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^2}=\frac{1}{4}\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^2}$$$$+\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}\Rightarrow$$$$\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}=\frac{3}{4}\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}$$$$\Rightarrow \sum _{p=1}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}=\frac{{\pi }^2}{8}-1\Rightarrow$$$$A=1-\frac{{\pi }^2}{8}$$$$$$

Commented by mnjuly1970 last updated on 19/Sep/20

$$thankyou.verynice..$$

Commented by mathmax by abdo last updated on 19/Sep/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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