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Question Number 74861 by aliesam last updated on 02/Dec/19

Answered by mind is power last updated on 02/Dec/19

$${\left({\ln }^2\left(\mathrm{x}\right)\right)}^{\frac{1}{2}}=-\ln \left(\mathrm{x}\right)$$$${\left(\ln \left({\mathrm{x}}^2\right)\right)}^{\frac{-1}{2}}=\frac{1}{\sqrt{2}}.{\left(\frac{1}{\ln \left(\mathrm{x}\right)}\right)}^{\frac{1}{2}}$$$${\left({\ln }^2\left(\mathrm{x}\right)\right)}^{\frac{1}{2}}.{\left(\ln \left({\mathrm{x}}^2\right)\right)}^{-\frac{1}{2}}+\sqrt{\ln \left(\mathrm{x}\right)}$$$$=-\frac{\sqrt{\ln \left(\mathrm{x}\right)}}{\sqrt{2}}+\sqrt{\ln \left(\mathrm{x}\right)}=\frac{\sqrt{2}-1}{\sqrt{2}}\sqrt{\ln \left(\mathrm{x}\right)}$$$$\frac{\ln \left(\mathrm{x}\right)+1}{\sqrt{2\ln \left(\mathrm{x}\right)}}-\sqrt{\ln \left(\sqrt{\mathrm{x}}\right))}$$$$=\frac{\ln \left(\mathrm{x}\right)+1}{\sqrt{2\ln \left(\mathrm{x}\right)}}-\sqrt{\left(\frac{\ln \left(\mathrm{x}\right)}{2}\right)}$$$$=\frac{\sqrt{\ln \left(\mathrm{x}\right)}}{\sqrt{2}}+\frac{1}{\sqrt{2\ln \left(\mathrm{x}\right)}}-\sqrt{\frac{\ln \left(\mathrm{x}\right)}{2}}$$$$=\sqrt{\frac{\ln \left(\mathrm{x}\right)}{2}}+\frac{1}{\sqrt{2\ln \left(\mathrm{x}\right)}}-\sqrt{\frac{\ln \left(\mathrm{x}\right)}{2}}=\frac{1}{\sqrt{2\ln \left(\mathrm{x}\right)}}$$$$\left(\frac{{\ln }^2\left(\mathrm{x}\right){\left(\ln \left({\mathrm{x}}^2\right)\right)}^{-\frac{1}{2}}+\sqrt{\ln \left(\mathrm{x}\right)}}{\frac{\ln \left(\mathrm{x}\right)+1}{\sqrt{2\ln \left(\mathrm{x}\right)}}-\sqrt{\ln \left(\sqrt{\mathrm{x}}\right))}}\right)=\frac{\sqrt{2}-1}{\sqrt{2}}\ln \left(\mathrm{x}\right).\frac{1}{\frac{1}{\sqrt{2\ln \left(\mathrm{x}\right)}}}=(\sqrt{2}-1)\ln \left(\mathrm{x}\right)$$$${\int }_0^1(\sqrt{2}-1)\ln \left(\mathrm{x}\right)\mathrm{dx}=(\sqrt{2}-1){\int }_0^1\ln \left(\mathrm{x}\right)\mathrm{dx}$$$$=(\sqrt{2}-1){[\mathrm{xln}\left(\mathrm{x}\right)-\mathrm{x}]}_0^1=-(\sqrt{2}-1)$$

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