5.5 The Substitution Rule/69: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 2: | Line 2: | ||
\begin{align} | \begin{align} | ||
\int_{0}^{1} \left(\frac{e^z + 1}{e^z + z}\right) &= \int_{}^{} \left((e^z +1) (\frac{1}{e^z +z}) \right) | \int_{0}^{1} \left(\frac{e^z + 1}{e^z + z}\right) &= \int_{}^{} \left((e^z +1) (\frac{1}{e^z +z}) \right) | ||
u= e^z + z | u= e^z + z | ||
Revision as of 15:47, 6 September 2022
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_{}^{} \left((e^z +1) (\frac{1}{e^z +z}) \right) u= e^z + z du= e^z +1 \end{align} }