int frac{1}{xsqrt{1 + log x}} , dx = | vedclass

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Question

(C) Let $t = 1 + \log x$. Then,the derivative is $dt = \frac{1}{x} \, dx$. Substituting these into the integral: $\int \frac{1}{\sqrt{t}} \, dt = \int

Vedclass · Accepted Answer

$2\sqrt{1 + \log x} + c$

Vedclass · Answer

(C) Let $t = 1 + \log x$. Then,the derivative is $dt = \frac{1}{x} \, dx$. Substituting these into the integral: $\int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt$. Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$: $= \frac{t^{1/2}}{1/2} + c = 2t^{1/2} + c$. Substituting $t = 1 + \log x$ back,we get: $= 2\sqrt{1 + \log x} + c$.

$\int \frac{1}{x\sqrt{1 + \log x}} \, dx = $

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