∫ 1 2 x x − 1 d x {\displaystyle \int _{1}^{2}x{\sqrt {x-1}}dx}
u = x − 1 u + 1 = x d u = 1 d x d u = d x {\displaystyle {\begin{aligned}u&=x-1\\u+1&=x\\[2ex]du&=1dx\\[2ex]du&=dx\end{aligned}}}
∫ 1 2 ( x x − 1 ) d x = ∫ 0 1 ( ( u + 1 ) u ) d u = ∫ 0 1 ( u 3 2 + u 1 2 ) d u = ( 2 5 u 5 2 + 2 3 u 3 2 ) | 0 1 = 2 5 + 2 3 = 16 15 {\displaystyle {\begin{aligned}\int _{1}^{2}(x{\sqrt {x-1}}\,)\;dx&=\int _{0}^{1}((u+1){\sqrt {u}})\;du=\int _{0}^{1}(u^{\frac {3}{2}}+u^{\frac {1}{2}})\;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}}}
![{\displaystyle {\begin{aligned}u&=x-1\\u+1&=x\\[2ex]du&=1dx\\[2ex]du&=dx\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e8a8c63930382b06bb5c51424e1ce72b64946c60)
![{\displaystyle {\begin{aligned}\int _{1}^{2}(x{\sqrt {x-1}}\,)\;dx&=\int _{0}^{1}((u+1){\sqrt {u}})\;du=\int _{0}^{1}(u^{\frac {3}{2}}+u^{\frac {1}{2}})\;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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8e29739b13b3ec4bdacccfb6a585fa103bcf83b9)