∫ 0 1 ( e z + 1 e z + z ) {\displaystyle \int _{0}^{1}\left({\frac {e^{z}+1}{e^{z}+z}}\right)}
u = e z + z d u = ( e z + 1 ) d x {\displaystyle {\begin{aligned}u&=e^{z}+z\\[2ex]du&=(e^{z}+1)dx\\[2ex]\end{aligned}}}
New upper limit: 1 = e 1 + 1 = e + 1 {\displaystyle 1=e^{1}+1=e+1} New lower limit: 0 = e 0 + 0 = 1 {\displaystyle 0=e^{0}+0=1}
∫ 0 1 ( e z + 1 e z + z ) = ∫ 0 1 ( ( e z + 1 ) d x ( 1 e z + z ) ) = ∫ 1 e + 1 ( 1 u ) d u = ( ln ( | u | ) ) {\displaystyle {\begin{aligned}\int _{0}^{1}\left({\frac {e^{z}+1}{e^{z}+z}}\right)&=\int _{0}^{1}\left((e^{z}+1)dx({\frac {1}{e^{z}+z}})\right)&=\int _{1}^{e+1}\left({\frac {1}{u}}\right)du&=\left(\ln(|u|)\right)\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=e^{z}+z\\[2ex]du&=(e^{z}+1)dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/88d16614472a3960a39988c31fd5c7ab6cc6c295)
