∫ 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 | ) ) | 1 e + 1 = ln ( | e + 1 | ) − ln ( | 1 | ) = ln ( e + 1 ) − 0 = ln ( e + 1 ) {\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)\\[2ex]&=\int _{1}^{e+1}\left({\frac {1}{u}}\right)du\\[2ex]&=\left(\ln(|u|)\right){\bigg |}_{1}^{e+1}\\[2ex]&=\ln(|e+1|)-\ln(|1|)\\[2ex]&=\ln(e+1)-0=\ln(e+1)\end{aligned}}}