Question-74861

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

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

{(\ln^{2} (x))}^{\frac{1}{2}} = - \ln (x)

{(\ln (x^{2}))}^{\frac{- 1}{2}} = \frac{1}{\sqrt{2}} . {(\frac{1}{\ln (x)})}^{\frac{1}{2}}

{(\ln^{2} (x))}^{\frac{1}{2}} . {(\ln (x^{2}))}^{- \frac{1}{2}} + \sqrt{\ln (x)}

= - \frac{\sqrt{\ln (x)}}{\sqrt{2}} + \sqrt{\ln (x)} = \frac{\sqrt{2} - 1}{\sqrt{2}} \sqrt{\ln (x)}

\frac{\ln (x) + 1}{\sqrt{2 \ln (x)}} - \sqrt{\ln (\sqrt{x}))}

= \frac{\ln (x) + 1}{\sqrt{2 \ln (x)}} - \sqrt{(\frac{\ln (x)}{2})}

= \frac{\sqrt{\ln (x)}}{\sqrt{2}} + \frac{1}{\sqrt{2 \ln (x)}} - \sqrt{\frac{\ln (x)}{2}}

= \sqrt{\frac{\ln (x)}{2}} + \frac{1}{\sqrt{2 \ln (x)}} - \sqrt{\frac{\ln (x)}{2}} = \frac{1}{\sqrt{2 \ln (x)}}

(\frac{\ln^{2} (x) {(\ln (x^{2}))}^{- \frac{1}{2}} + \sqrt{\ln (x)}}{\frac{\ln (x) + 1}{\sqrt{2 \ln (x)}} - \sqrt{\ln (\sqrt{x}))}}) = \frac{\sqrt{2} - 1}{\sqrt{2}} \ln (x) . \frac{1}{\frac{1}{\sqrt{2 \ln (x)}}} = (\sqrt{2} - 1) \ln (x)

\int_{0}^{1} (\sqrt{2} - 1) \ln (x) dx = (\sqrt{2} - 1) \int_{0}^{1} \ln (x) dx

= (\sqrt{2} - 1) {[xln (x) - x]}_{0}^{1} = - (\sqrt{2} - 1)