∫ 0 1 x 2 ( 1 + 2 x 3 ) 5 d r {\displaystyle \int _{0}^{1}x^{2}(1+2x^{3})^{5}dr}
u = 1 + 2 x 3 {\displaystyle {\begin{aligned}u&=1+2x^{3}\\[2ex]\end{aligned}}}
d u = 6 x 2 d x {\displaystyle {\begin{aligned}du&=6x^{2}dx\\[2ex]\end{aligned}}}
∫ 1 3 u 5 1 6 d u {\displaystyle {\begin{aligned}\int _{1}^{3}u^{5}{\frac {1}{6}}du\\[2ex]\end{aligned}}}
1 36 ( 1 + 2 x 3 ) 6 < m a t h > = 1 6 ∫ 1 3 u 5 d u {\displaystyle {\begin{aligned}{\frac {1}{36}}(1+2x^{3})^{6}\end{aligned}}<math>{\begin{aligned}&={\frac {1}{6}}\int _{1}^{3}u^{5}du\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=1+2x^{3}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/bb7af9540f66d41bbcd61431c87e360f7975b4a3)
![{\displaystyle {\begin{aligned}du&=6x^{2}dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/62e39f21f3ca2c85aa04408ea0ace6e63450da2b)
![{\displaystyle {\begin{aligned}\int _{1}^{3}u^{5}{\frac {1}{6}}du\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/4d064cc17ebb62890cfd46e31e5da7ddf82e9cc1)
![{\displaystyle {\begin{aligned}{\frac {1}{36}}(1+2x^{3})^{6}\end{aligned}}<math>{\begin{aligned}&={\frac {1}{6}}\int _{1}^{3}u^{5}du\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6099345b6a53c5b7b162b56d24e27aa4735094f8)