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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{1}^{e+1} \left((e^z +1) (\frac{1}{e^z +z}) \right) }