5.5 The Substitution Rule/69

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{1}\left({\frac {e^{z}+1}{e^{z}+z}}\right)}

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}u&=e^{z}+z\\[2ex]du&=(e^{z}+1)dx\\[2ex]\end{aligned}}}

New upper limit: 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 1 = e^1 + 1 = e + 1 }
New lower limit: 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 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_{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{align} }