int frac{1}{x} sec^2(log x) , dx = | vedclass

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Question

(A) Let $t = \log x$. Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $dt = \frac{1}{x} \, dx$.

Vedclass · Accepted Answer

$	an(\log x) + c$

Vedclass · Answer

(A) Let $t = \log x$. Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $dt = \frac{1}{x} \, dx$. Substituting these into the integral: $\int \sec^2(\log x) \cdot \frac{1}{x} \, dx = \int \sec^2(t) \, dt$. The integral of $\sec^2(t)$ is $	an(t) + c$. Replacing $t$ with $\log x$,we get $	an(\log x) + c$.

$\int \frac{1}{x} \sec^2(\log x) \, dx = $

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