5.5 The Substitution Rule/65

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

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}u&=/xx-1\\[2ex]du&={\frac {1}{2}}\ {\frac {1}{\sqrt {t}}}dx\\[2ex]2du&={\frac {1}{\sqrt {t}}}dx\end{aligned}}}