∫ 0 1 ( x 2 + 1 ) e − x d x {\displaystyle \int _{0}^{1}\left(x^{2}+1\right)e^{-x}dx}
u = x 2 + 1 d v = e − x d x d u = 2 x d x v = − e − x {\displaystyle {\begin{aligned}u&=x^{2}+1\quad dv=e^{-x}dx\\[2ex]du&=2xdx\qquad v=-e^{-x}\\[2ex]\end{aligned}}}
= ( x 2 + 1 ) ( − e − x ) | 0 1 − 2 ∫ 0 1 ( − e x ) ( x ) d x {\displaystyle {\begin{aligned}&=(x^{2}+1)(-e^{-x}){\bigg |}_{0}^{1}-2\int _{0}^{1}(-e^{x})(x)dx\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=x^{2}+1\quad dv=e^{-x}dx\\[2ex]du&=2xdx\qquad v=-e^{-x}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dca2a9cf6138c41cb61d56c1ee91c9c4fea73909)
![{\displaystyle {\begin{aligned}&=(x^{2}+1)(-e^{-x}){\bigg |}_{0}^{1}-2\int _{0}^{1}(-e^{x})(x)dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/06474884d0ba6d6fc8fae829d72953453d786c3a)