Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 43

$$\int x(x-1)^{10}dx=\frac{(x-1)^{12}}{12}+\frac{(x-1)^{11}}{11}+C$$

$$A=\int x(x-1)^{10}dx$$ We set $u=x-1$, which means $x=u+1$ Then $$du=dx$$ Therefore, $$A=\int(u+1)u^{10}du=\int\Big(u^{11}+u^{10}\Big)du$$ $$A=\frac{u^{12}}{12}+\frac{u^{11}}{11}+C$$ $$A=\frac{(x-1)^{12}}{12}+\frac{(x-1)^{11}}{11}+C$$