Evaluate the integral from 0 to 1 of (x^3 - x^2)/log(x) dx.

Evaluate the integral from 0 to 1 of (x^3 - x^2)/log(x) dx.

Approach: Introduce a parameter in one exponent, differentiate under the integral sign to kill the logarithm, integrate the resulting elementary function, then fix the constant from a value of the parameter where the integral is visibly zero.

log(4/3). Define I(b) as the integral from 0 to 1 of (x^b - x^2)/log(x) dx for b > -1. Differentiation under the integral sign with respect to b gives I'(b) = the integral from 0 to 1 of x^b * log(x)/log(x) dx = the integral of x^b dx = 1/(b+1), which is where the logarithm cancels. Integrating gives I(b) = log(b+1) + C, and the boundary condition I(2) = 0, since the integrand vanishes identically at b = 2, forces C = -log(3). So I(b) = log((b+1)/3) and I(3) = log(4/3) = 0.2877. The interchange is legitimate because the integrand and its b derivative are continuous and dominated on [0, 1] for b in a neighbourhood of the values used, with the apparent singularity at x = 1 removable since (x^b - x^2)/log(x) tends to b - 2 there.

Follow-up: Evaluate the same integral with x^3 - x^2 replaced by sin(log(1/x)) times a power of x, and say where the parameter should go?

Key concepts: differentiation under the integral sign, parameter integral, logarithm, boundary condition.