int left[ frac{1+log x}{cos^{2}(x log x)} rig | vedclass

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Question

(D) Let $I = \int \frac{1+\log x}{\cos^{2}(x \log x)} dx$.Substitute $t = x \log x$.Differentiating with respect to $x$ using the product rule: $\frac

Vedclass · Accepted Answer

$	an(x \log x) + c$

Vedclass · Answer

(D) Let $I = \int \frac{1+\log x}{\cos^{2}(x \log x)} dx$.Substitute $t = x \log x$.Differentiating with respect to $x$ using the product rule: $\frac{dt}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x$.Therefore,$(1 + \log x) dx = dt$.Substituting these into the integral: $I = \int \frac{1}{\cos^{2} t} dt = \int \sec^{2} t dt$.The integral of $\sec^{2} t$ is $	an t + c$.Substituting back $t = x \log x$,we get $I = 	an(x \log x) + c$.

$\int \left[ \frac{1+\log x}{\cos^{2}(x \log x)} \right] dx =$

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