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: $0=e^{0}+0=1$

Failed to parse (unknown function "\abs"): {\displaystyle \begin{align} \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) |bigg \end{align} }