$\displaystyle I = \displaystyle \int_{}{} \frac{dx}{x\sqrt{1-(In x)^2}}$ help
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Let $\displaystyle u = \ln{x}$ and it becomes $\displaystyle \int\frac{1}{\sqrt{1-u^2}}\;{du}$ (a standard integral).
Originally Posted by Hardwork Let $\displaystyle u = \ln{x}$ and it becomes $\displaystyle \int\frac{1}{\sqrt{1-u^2}}\;{du}$ (a standard integral). thanks, I feel stupid for missing how obvious it is
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