∫ x 2 x 3 + 1 d x {\displaystyle \int x^{2}{\sqrt {x^{3}+1}}dx}
u = x 3 + 1 d u = 3 x 2 d x {\displaystyle {\begin{aligned}u&=x^{3}+1\\[2ex]du&=3x^{2}dx\end{aligned}}}
∫ 1 3 d u u {\displaystyle \int {\frac {1}{3}}du{\sqrt {u}}}
1 3 ∫ d u ( u 1 2 {\displaystyle {\frac {1}{3}}\int du(u^{\frac {1}{2}}}