5.5 The Substitution Rule/69: Difference between revisions

Revision as of 15:52, 6 September 2022

$\int _{0}^{1}\left({\frac {e^{z}+1}{e^{z}+z}}\right)$

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} u &= e^z + z \\[2ex] du &= e^z +1 \\[2ex] /end{align} }

New upper limit: Failed to parse (syntax error): {\displaystyle \1 = e^1 + 1 = e + 1 }
New lower limit: $0=e^{0}+0=1$

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int_{1}^{e+1} \left((e^z +1) (\frac{1}{e^z +z}) \right) }

@@ Line 1: / Line 1: @@
 <math>
+\int_{0}^{1} \left(\frac{e^z + 1}{e^z + z}\right)
+</math>
+<math>
 \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)
-\end{align}
+u &= e^z + z \\[2ex]
-</math>
-<math>
+du &= e^z +1 \\[2ex]
-u= e^z + z
+/end{align}
+</math>
-du= e^z +1
+New upper limit: <math> \1 = e^1 + 1 = e + 1 </math><br>
+New lower limit: <math> 0 = e^0 + 0 = 1 </math>
+<math>
+\begin{align}
+\int_{1}^{e+1} \left((e^z +1) (\frac{1}{e^z +z}) \right)
 </math>