∫ 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}}}
∫ ( ln ( x ) n ) d x = x ln ( x ) n − ∫ ( ( x n ln ( x ) n − 1 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 _{}^{}\left(\ln(x)^{n}\right)dx&=x\ln(x)^{n}-\int _{}^{}\left((x{\frac {n\ln(x)^{n-1}}{x}})\right)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/dca2a9cf6138c41cb61d56c1ee91c9c4fea73909)
![{\displaystyle {\begin{aligned}\int _{}^{}\left(\ln(x)^{n}\right)dx&=x\ln(x)^{n}-\int _{}^{}\left((x{\frac {n\ln(x)^{n-1}}{x}})\right)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/19be1d5890fa5a4661b92c41d3728d2463eb1083)