\begin{frame}
  \frametitle{The Substitution Rule for Indefinite Integrals}

  \begin{exampleblock}{}
    \vspace{-.5ex}
    \begin{talign}
      \int \frac{x}{\sqrt{1-4x^2}} dx 
    \end{talign}
    \pause
    We choose $u = \pause 1-4x^2$. \pause Then $u' = \pause -8x$\pause, and hence
    \begin{talign}
      \int \frac{x}{\sqrt{1-4x^2}} dx 
      \mpause[1]{&= \int \frac{x}{\sqrt{u}}\, \frac{du}{-8x} }
      \mpause{= -\frac{1}{8} \int \frac{1}{\sqrt{u}}\, du }\\
      \mpause{&= -\frac{1}{8} \cdot 2\sqrt{u} + C}
      \mpause{= -\frac{1}{4} \sqrt{1-4x^2} + C}
    \end{talign}
  \end{exampleblock}
  \pause\pause\pause\pause\pause
  
  \begin{exampleblock}{}
    \vspace{-.5ex}
    \begin{talign}
      \int e^{5x} dx 
    \end{talign}
    \pause
    We choose $u = \pause 5x$. \pause Then $u' = \pause 5$\pause, and hence
    \begin{talign}
      \int e^{5x} dx 
      \mpause[1]{&= \int e^u\, \frac{du}{5} }
      \mpause{= \frac{1}{5} \int e^u\, du }\\
      \mpause{&= \frac{1}{5} e^u + C}
      \mpause{=  \frac{1}{5} e^{5x} + C}
    \end{talign}
  \end{exampleblock}
\end{frame}