\begin{frame}
  \frametitle{The Substitution Rule for Indefinite Integrals}
  \begin{malign}
    \int \sqrt{2x+1} dx 
  \end{malign}
  \pause\medskip
  
  \begin{exampleblock}{}
    We choose $u = \pause 2x+1$. \pause Then $u' = 2$\pause, and hence
    \begin{talign}
      \int \sqrt{2x+1} dx 
      \mpause[1]{&= \int \sqrt{u} \frac{du}{2} }
      \mpause{= \frac{1}{2} \int \sqrt{u}\, du }\\
      \mpause{&= \frac{1}{2} \cdot \frac{2}{3}u^{\frac{3}{2}} + C}
      \mpause{= \frac{1}{3} (2x+1)^{\frac{3}{2}} + C}
    \end{talign}
  \end{exampleblock}
  \pause\pause\pause\pause\pause\medskip
  
  We could have chosen another $u$. For example:
  \pause
  \begin{exampleblock}{}
    We choose $u = \sqrt{2x+1}$. \pause Then $u' = \pause \frac{1}{2\sqrt{2x+1}}\cdot 2 \pause= \frac{1}{u}$\pause. Then
    \begin{talign}
      \int \sqrt{2x+1} dx 
      \mpause[1]{&= \int u\, \frac{du}{1/u} }
      \mpause{= \int u^2\, du }\\
      \mpause{&= \frac{1}{3}u^{3} + C}
      \mpause{= \frac{1}{3}\big(\sqrt{2x+1}\big)^{3} + C}
      \mpause{= \frac{1}{3} (2x+1)^{\frac{3}{2}} + C}
    \end{talign}
  \end{exampleblock}
  \vspace{10cm}
\end{frame}