Standard Logarithmic Integrals

Integrating \ln x on its own relies on a clever device: the logarithm has no obvious antiderivative until we write it as the product 1\cdot\ln x and apply integration by parts. The underlying principle is that introducing an implicit factor of one supplies a dv we can integrate while we differentiate the troublesome logarithm away. Choosing u = \ln x and dv = dx, we get du = \frac{1}{x},dx and v = x. The parts formula \int u,dv = uv - \int v,du then yields x\ln x - \int x\cdot\frac{1}{x},dx = x\ln x - \int 1,dx = x\ln x - x + C. Option x\ln x + x makes a sign error on the second term. Option (\ln x)^2/2 incorrectly treats \ln x as if it were a simple power variable. Option 1/x confuses integrating the logarithm with differentiating it. The trick of inserting a factor of one as dv is a recurring JEE Advanced device worth memorizing. As a final plausibility check, differentiating x\ln x - x gives \ln x + x\cdot\frac{1}{x} - 1 = \ln x, exactly the integrand, valid for x > 0 where the logarithm is defined.