$\displaystyle I = \displaystyle \int_{}{} \frac{dx}{x\sqrt{1-(In x)^2}}$

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- Apr 9th 2011, 03:27 AMbijosnEValuate this indefinite integral?
$\displaystyle I = \displaystyle \int_{}{} \frac{dx}{x\sqrt{1-(In x)^2}}$

help - Apr 9th 2011, 03:36 AMHardwork
Let $\displaystyle u = \ln{x}$ and it becomes $\displaystyle \int\frac{1}{\sqrt{1-u^2}}\;{du}$ (a standard integral).

- Apr 9th 2011, 03:46 AMbijosn