5.5 The Substitution Rule/65

$\int _{1}^{2}x{\sqrt {x-1}}dx=\int _{0}^{1}u+1{\sqrt {u}}du=\int _{0}^{1}(u+1)({\sqrt {u}})=\int _{0}^{1}u^{\frac {2}{3}}+{\sqrt {u}}du={\frac {2}{5}}+{\frac {2}{3}}={\frac {16}{15}}$