∫ 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}}}
∫ 0 1 ( x 2 + 1 ) e − x d x = ( x 2 + 1 ) ( − e − x ) | 0 1 − 2 ∫ 0 1 ( − e x ) ( x ) d x = x ln ( x ) n − ∫ ( n ln ( x ) n − 1 ) d x = x ln ( x ) n − n ∫ ( ln ( x ) n − 1 ) d x {\displaystyle {\begin{aligned}\int _{0}^{1}\left(x^{2}+1\right)e^{-x}dx&=(x^{2}+1)(-e^{-x}){\bigg |}_{0}^{1}-2\int _{0}^{1}(-e^{x})(x)dx\\[2ex]&=x\ln(x)^{n}-\int _{}^{}\left(n\ln(x)^{n-1}\right)dx\\[2ex]&=x\ln(x)^{n}-n\int _{}^{}\left(\ln(x)^{n-1}\right)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/b0cca040a57dc8ca17171495ea48c0dfa3a1c508)
![{\displaystyle {\begin{aligned}\int _{0}^{1}\left(x^{2}+1\right)e^{-x}dx&=(x^{2}+1)(-e^{-x}){\bigg |}_{0}^{1}-2\int _{0}^{1}(-e^{x})(x)dx\\[2ex]&=x\ln(x)^{n}-\int _{}^{}\left(n\ln(x)^{n-1}\right)dx\\[2ex]&=x\ln(x)^{n}-n\int _{}^{}\left(\ln(x)^{n-1}\right)dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5d44df4e207b17487fadd9bdf1374507a2104612)