5.5 The Substitution Rule/69

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

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

New upper limit: Failed to parse (syntax error): {\displaystyle \1 = e^1 + 1 = e + 1 }
New lower limit: $0=e^{0}+0=1$

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int_{1}^{e+1} \left((e^z +1) (\frac{1}{e^z +z}) \right) }