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 17: / Line 17: @@
 <math>
-\sqrt{u}du=\frac{2u^\frac{2}{3}}{3}+C
+\sqrt{u}du=\frac{2u^{2/3}}{3}+C
 </math>
 <math>
-=\frac{2}{3}(e^x+1)^\frac{3}{2}
+=\frac{2}{3}(e^x+1)^{3/2}
 </math>