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Question Number 144764 by nonh1 last updated on 29/Jun/21

Answered by liberty last updated on 29/Jun/21

$${\int }_0^{\sqrt{\ln 2}}{\int }_0^1\frac{1}{2}{\mathrm{xe}}^{{\mathrm{x}}^2}\left(\frac{\mathrm{d}(1+{\mathrm{y}}^2)}{1+{\mathrm{y}}^2}\right)\mathrm{dx}$$$$={\int }_0^{\sqrt{\ln 2}}\frac{1}{2}{\mathrm{xe}}^{{\mathrm{x}}^2}\left(\ln (1+{\mathrm{y}}^2){\mid }_0^1\right)\mathrm{dx}$$$$=\frac{1}{4}\ln 2{\int }_0^{\sqrt{\ln 2}}{\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{d}\left({\mathrm{x}}^2\right)$$$$=\frac{1}{4}\ln 2{\left[{\mathrm{e}}^{{\mathrm{x}}^2}\right]}_0^{\sqrt{\ln 2}}$$$$=\frac{1}{4}\ln 2[{\mathrm{e}}^{\ln 2}-1]$$$$=\frac{1}{4}\ln 2[2-1]=\frac{1}{4}\ln 2$$

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