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Question Number 144997 by qaz last updated on 01/Jul/21

$${\int }_0^1\frac{\ln (1+\mathrm{x})\ln (1+{\mathrm{x}}^2)}{\mathrm{x}}\mathrm{dx}=\frac{\pi }{2}\mathrm{G}-\frac{33}{32}\zeta \left(3\right)$$

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