∫ 1 2 x x − 1 d x = ∫ 0 1 u + 1 u d u = ∫ 0 1 ( u + 1 ) ( u ) = ∫ 0 1 u 3 2 + u d u = 2 5 U 5 2 + 2 3 U 3 2 | 0 1 = 2 5 + 2 3 = 16 15 {\displaystyle \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}}}
u = x − 1 d u = 2 d x 2 d u = 1 t d x {\displaystyle {\begin{aligned}u&=x-1\\[2ex]du&=2dx\\[2ex]2du&={\frac {1}{\sqrt {t}}}dx\end{aligned}}}