5.5 The Substitution Rule/25: Difference between revisions

Latest revision as of 19:10, 2 September 2022

$\int e^{x}{\sqrt {1+e^{x}}}dx$

$u=1+e^{x}$

$du=e^{x}dx$

$\int e^{x}{\sqrt {1+e^{x}}}dx=\int {\sqrt {u}}du$

${\sqrt {u}}du={\frac {2u^{2/3}}{3}}+C$

$={\frac {2}{3}}(e^{x}+1)^{3/2}$

@@ Line 1: / Line 1: @@
 <math>
 \int e^x\sqrt{1+e^x}dx
+</math>
+<math>
+u=1+e^x
+</math>
+<math>
+du=e^xdx
+</math>
+<math>
+\int e^x\sqrt{1+e^x}dx=\int\sqrt{u}du
+</math>
+<math>
+\sqrt{u}du=\frac{2u^{2/3}}{3}+C
+</math>
+<math>
+=\frac{2}{3}(e^x+1)^{3/2}
 </math>