∫ 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 6 ∫ 1 3 u 5 d u {\displaystyle {\begin{aligned}&={\frac {1}{6}}\int _{1}^{3}u^{5}du\\[2ex]\end{aligned}}}
= 1 36 ( 1 + 2 x 3 ) 6 {\displaystyle {\begin{aligned}&={\frac {1}{36}}(1+2x^{3})^{6}\end{aligned}}}
= 181 9 {\displaystyle {\begin{aligned}&={\frac {181}{9}}\\[2ex]\end{aligned}}}