5.5 The Substitution Rule/69

$\int _{0}^{1}\left({\frac {e^{z}+1}{e^{z}+z}}\right)$

${\begin{aligned}u&=e^{z}+z\\[2ex]du&=(e^{z}+1)dx\\[2ex]\end{aligned}}$

New upper limit: $1=e^{1}+1=e+1$
New lower limit: $0=e^{0}+0=1$

${\begin{aligned}\int _{0}^{1}\left({\frac {e^{z}+1}{e^{z}+z}}\right)&=\int _{0}^{1}\left((e^{z}+1)dx({\frac {1}{e^{z}+z}})\right)&=\int _{1}^{e+1}\left({\frac {1}{u}}\right)du&=\left(\ln(abs(u))\right)\end{aligned}}$