5.5 The Substitution Rule/69

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 \int_{0}^{1} \left(\frac{e^z + 1}{e^z + z}\right) }

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

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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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)\\[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|)&=\ln(e+1)-0=\ln(e+1)\end{aligned}}}