Prove ∫ ( x n e x ) d x = x n e x − n ∫ ( x n − 1 e x ) d x {\displaystyle \int _{}^{}\left(x^{n}e^{x}\right)dx=x^{n}e^{x}-n\int _{}^{}\left(x^{n-1}e^{x}\right)dx}
u = x n d v = e x d x d u = n x n − 1 d x v = e x {\displaystyle {\begin{aligned}u&=x^{n}\quad &dv=e^{x}dx\\[2ex]du&=nx^{n-1}dx\quad &v=e^{x}\\[2ex]\end{aligned}}}
∫ ( x n e x ) d x = x n e x − ∫ ( n x n − 1 e x ) d x = x n e x − n ∫ ( x n − 1 e x ) d x {\displaystyle {\begin{aligned}\int _{}^{}\left(x^{n}e^{x}\right)dx&=x^{n}e^{x}-\int _{}^{}\left(nx^{n-1}e^{x}\right)dx\\[2ex]&=x^{n}e^{x}-n\int _{}^{}\left(x^{n-1}e^{x}\right)dx\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=x^{n}\quad &dv=e^{x}dx\\[2ex]du&=nx^{n-1}dx\quad &v=e^{x}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8cc421691d2243b55bbe0a0fd3e1babcb1049981)
![{\displaystyle {\begin{aligned}\int _{}^{}\left(x^{n}e^{x}\right)dx&=x^{n}e^{x}-\int _{}^{}\left(nx^{n-1}e^{x}\right)dx\\[2ex]&=x^{n}e^{x}-n\int _{}^{}\left(x^{n-1}e^{x}\right)dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b7548e0ad0dcf7796ae2f6e505141f27e2ea8399)