Thomas' Calculus 13th Edition

$6 \ln 2$
Integrating the inside $dy$ integral, we get: $I= \int_{2}^{\infty} [ \dfrac{3 (y-1)^{1/3} dx}{x^2-x}]_0^2$ or, $= \int_{2}^{\infty} \dfrac{6}{x^2-x} dx$ or, $= 6 \int_{2}^{\infty} [\dfrac{1}{x-1} -\dfrac{1}{x}]dx$ or, $= 6 \lim\limits_{b \to \infty } \int_{2}^{b} [\dfrac{1}{x-1} -\dfrac{1}{x}]dx$ or, $= 6 \lim\limits_{b \to \infty } \ln (b-1) -\ln b -\ln 1+\ln 2$ Thus, we have $I=6 \lim\limits_{b \to \infty } \ln (1-\dfrac{1}{b})+\ln (2)=6 \ln 2$