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

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Question

(A) Let $I = \int \frac{1}{x \cos^2(1 + \log x)} \, dx$. Substitute $t = 1 + \log x$. Then,differentiating both sides with respect to $x$,we get $\fra

Vedclass · Accepted Answer

$	an(1 + \log x) + c$

Vedclass · Answer

(A) Let $I = \int \frac{1}{x \cos^2(1 + \log x)} \, dx$. Substitute $t = 1 + \log x$. 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,we get: $I = \int \frac{1}{\cos^2 t} \, dt = \int \sec^2 t \, dt$. Since the integral of $\sec^2 t$ is $	an t + c$,we have: $I = 	an t + c$. Substituting back $t = 1 + \log x$,we get: $I = 	an(1 + \log x) + c$.

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

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