int frac{tan(log x)}{x} , dx = | vedclass

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Question

(C) Let $I = \int \frac{	an(\log x)}{x} \, dx$. Substitute $\log x = t$. Then,differentiating with respect to $x$,we get $\frac{1}{x} \, dx = dt$. Su

Vedclass · Accepted Answer

$\log \sec(\log x) + c$

Vedclass · Answer

(C) Let $I = \int \frac{	an(\log x)}{x} \, dx$. Substitute $\log x = t$. Then,differentiating with respect to $x$,we get $\frac{1}{x} \, dx = dt$. Substituting these into the integral,we get: $I = \int 	an(t) \, dt$. The integral of $	an(t)$ is $\log |\sec(t)| + c$. Substituting $t = \log x$ back,we get: $I = \log |\sec(\log x)| + c$.

$\int \frac{\tan(\log x)}{x} \, dx = $

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