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

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