∫ x 3 ( 2 + x 4 ) 5 d x , u = 2 + x 4 {\displaystyle \int x^{3}(2+x^{4})^{5}dx{\text{,}}\quad u=2+x^{4}}
u = 2 + x 4 d u = 4 x 3 d x 1 4 d u = x 3 d x {\displaystyle {\begin{aligned}u&=2+x^{4}\\[2ex]du&=4x^{3}dx\\[2ex]{\frac {1}{4}}du&=x^{3}dx\end{aligned}}}
∫ x 3 ( 2 + x 4 ) 5 d x = ∫ ( x 3 d x ) ( 2 + x 4 ) = ∫ ( 1 4 d u ) ( u ) = 1 4 ∫ u d u = 1 4 [ u 1 + 1 1 + 1 ] + C = u 2 8 + C = ( 2 + x 4 ) 2 8 + C {\displaystyle {\begin{aligned}\int x^{3}(2+x^{4})^{5}dx&=\int (x^{3}dx)(2+x^{4})=\int \left({\frac {1}{4}}du\right)(u)={\frac {1}{4}}\int u\,du\\[2ex]&={\frac {1}{4}}\left[{\frac {u^{1+1}}{1+1}}\right]+C={\frac {u^{2}}{8}}+C\\[2ex]&={\frac {(2+x^{4})^{2}}{8}}+C\end{aligned}}}