5.5 The Substitution Rule/65

$\int _{1}^{2}x{\sqrt {x-1}}dx$

${\begin{aligned}u&=x-1\\u+1&=x\\[2ex]du&=1dx\\[2ex]du&=dx\end{aligned}}$

${\begin{aligned}\int _{1}^{2}x{\sqrt {x-1}}\,dx&=\int _{0}^{1}(u+1){\sqrt {u}}\,du=\int _{0}^{1}(u+1)({\sqrt {u}})=\int _{0}^{1}u^{\frac {3}{2}}+{\sqrt {u}}du\\[2ex]&={\frac {2}{5}}(u^{\frac {5}{2}}+{\frac {2}{3}}u^{\frac {3}{2}}{\bigg |})_{0}^{1}={\frac {2}{5}}+{\frac {2}{3}}\\[2ex]&={\frac {16}{15}}\\[2ex]\end{aligned}}$