int frac{dx}{xsqrt{1 - (log x)^2}} = | vedclass

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(C) Let $I = \int \frac{dx}{x\sqrt{1 - (\log x)^2}}$.Substitute $t = \log x$.Then,the derivative is $dt = \frac{1}{x} dx$.Substituting these into the

Vedclass · Accepted Answer

$\sin^{-1}(\log x) + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{x\sqrt{1 - (\log x)^2}}$.Substitute $t = \log x$.Then,the derivative is $dt = \frac{1}{x} dx$.Substituting these into the integral,we get:$I = \int \frac{dt}{\sqrt{1 - t^2}}$.Using the standard integral formula $\int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$,we obtain:$I = \sin^{-1}(t) + c$.Substituting back $t = \log x$,we get:$I = \sin^{-1}(\log x) + c$.

$\int \frac{dx}{x\sqrt{1 - (\log x)^2}} = $

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