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Question Number 142935 by mnjuly1970 last updated on 07/Jun/21

Answered by qaz last updated on 07/Jun/21

$${\int }_0^1\frac{{\ln }^2(1-{\mathrm{x}}^2)}{\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1\frac{{\ln }^2(1-{\mathrm{x}}^2)}{{\mathrm{x}}^2}\mathrm{d}\left({\mathrm{x}}^2\right)$$$$=\frac{1}{2}{\int }_0^1\frac{{\ln }^2\mathrm{u}}{1-\mathrm{u}}\mathrm{du}$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{u}}^{\mathrm{n}}{\ln }^2\mathrm{udu}$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{2}2!}{{(\mathrm{n}+1)}^3}$$$$=\zeta \left(3\right)$$

Commented by mnjuly1970 last updated on 07/Jun/21

$$thankyouverymuch...$$$$$$

Answered by Dwaipayan Shikari last updated on 07/Jun/21

$$x^2=u\Rightarrow dx=\frac{1}{2}.\frac{du}{x}$$$$=\frac{1}{2}{\int }_0^1\frac{{log}^2(1-u)}{u}du$$$$=\frac{1}{2}{\int }_0^1\frac{{log}^2\left(u\right)}{1-u}du=\frac{1}{2}.{\psi }^{″}\left(1\right)=\zeta \left(3\right)$$

Commented by mnjuly1970 last updated on 07/Jun/21

$$thanksalot....$$$$$$

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