Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 42

$$\int\sqrt{\frac{x^4}{x^3-1}}dx=\frac{2\sqrt{x^3-1}}{3}+C$$

$$A=\int\sqrt{\frac{x^4}{x^3-1}}dx=\int\frac{x^2}{\sqrt{x^3-1}}dx$$ We set $u=\sqrt{x^3-1}$ Then $$du=\frac{(x^3-1)'}{2\sqrt{x^3-1}}dx=\frac{3x^2}{2\sqrt{x^3-1}}dx$$ That means, $$\frac{x^2}{\sqrt{x^3-1}}dx=\frac{2}{3}du$$ Therefore, $$A=\frac{2}{3}\int du=\frac{2}{3}u+C$$ $$A=\frac{2\sqrt{x^3-1}}{3}+C$$