Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 43

$\frac{1}{12}(x-1)^{12}+\frac{1}{11}(x-1)^{11}+C$

Let u$=x-1.$ Then du=dx and x=u+1. Thus $\int x(x=1)^{10}dx$ =$\int (u+1)u^{10}du$ =$\int (u^{11}+u^{10})$ =$\frac{1}{12}u^{12}+\frac{1}{11}u^{11}+C$ =$\frac{1}{12}(x-1)^{12}+\frac{1}{11}(x-1)^{11}+C$