∫ 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 ) = 1 2 u 3 2 {\displaystyle {\begin{aligned}{\frac {1}{3}}\int du(u^{\frac {1}{2}})\\[2ex]={\frac {1}{2}}u^{\frac {3}{2}}\end{aligned}}}
1 2 ( x 3 + 1 ) 3 2 + C {\displaystyle {\frac {1}{2}}(x^{3}+1)^{\frac {3}{2}}+C}
![{\displaystyle {\begin{aligned}u&=x^{3}+1\\[2ex]du&=3x^{2}dx\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/c2f79f5349cf765f09cba734b277a92effa63f43)
![{\displaystyle {\begin{aligned}{\frac {1}{3}}\int du(u^{\frac {1}{2}})\\[2ex]={\frac {1}{2}}u^{\frac {3}{2}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/477a7d2591016b42cb7ef37b7733fe5fb5a99c31)
