Indefinite Integrals

The integrand \frac{1}{x \ln x} signals a substitution because the derivative of \ln x is exactly 1/x, which already appears as a multiplicative factor in the expression. The guiding principle is u-substitution, which reverses the chain rule: whenever a composite function sits beside its own inner derivative, we can collapse the whole expression onto a single new variable. Let u = \ln x, so that du = \frac{1}{x},dx. The integral transforms cleanly into \int \frac{du}{u}, a standard reciprocal form whose antiderivative is the logarithm \ln|u| + C. Back-substituting u = \ln x produces \ln|\ln x| + C. This problem rewards recognizing the pairing of a function with its derivative, a skill the JEE Advanced tests repeatedly. Option (\ln x)^2/2 wrongly treats \ln x as the variable being integrated linearly rather than reciprocally. Option \ln|x| ignores the inner logarithm entirely and integrates as though the denominator were just x. Option 1/\ln x would arise only if the integrand were -1/(x(\ln x)^2), which is a genuinely different problem. As a final plausibility check, differentiating \ln|\ln x| via the chain rule gives \frac{1}{\ln x}\cdot\frac{1}{x}, which is exactly the original integrand, confirming correctness across the valid domain x > 1 where the outer logarithm is defined.