∫ 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}}
Given u = 2 + x 4 then d u = 4 x 3 d x 1 4 d u = x 3 d x {\displaystyle {\begin{aligned}{\text{Given }}u&=2+x^{4}\\[2ex]{\text{then }}du&=4x^{3}dx\\[2ex]{\frac {1}{4}}du&=x^{3}dx\end{aligned}}}
d u = ( 4 x 3 ) d x {\displaystyle du=(4x^{3})\,dx} d u 4 = x 3 d x {\displaystyle {\frac {du}{4}}=x^{3}dx}
∫ x 3 ( 2 + x 4 ) 5 d x = ∫ {\displaystyle \int x^{3}(2+x^{4})^{5}dx=\int }
![{\displaystyle {\begin{aligned}{\text{Given }}u&=2+x^{4}\\[2ex]{\text{then }}du&=4x^{3}dx\\[2ex]{\frac {1}{4}}du&=x^{3}dx\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/f7e6d0d4ed39e680f8b31536025b1d8665ae6af1)
