5.5 The Substitution Rule/25: Difference between revisions

Revision as of 19:06, 2 September 2022

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

$u=1+e^{x}$

$du=e^{x}dx$

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

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

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

@@ Line 13: / Line 13: @@
 <math>
-\int\sqrt{u} du
+\int e^x\sqrt{1+e^x}dx=\int\sqrt{u}du
 </math>
 <math>
-\sqrt{u}du=\frac{2u^\frac{2}{3}}{3}
+\sqrt{u}du=\frac{2u^\frac{2}{3}}{3}+C
+</math>
+<math>
+=\frac{2}{3}(e^x+1)^\frac{3}{2}
 </math>