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

  A slightly more interesting example:
  \begin{exampleblock}{}
    \vspace{-.5ex}
    \begin{talign}
      \int x^5\sqrt{1+x^2}\, dx 
    \end{talign}
    \pause
    We choose $u = \pause 1+x^2$. \pause Then $u' = \pause 2x$\pause, and hence
    \begin{talign}
      \int x^5\sqrt{1+x^2}\, dx 
      \mpause[1]{&= \int x^5\sqrt{u}\, \frac{du}{2x} }
      \mpause{= \frac{1}{2} \int x^4\sqrt{u}\, du }
    \end{talign}
    \pause\pause\pause
    What now? \pause Note that
    \quad$x^2 = u-1$\quad and \quad$x^4 = (x^2)^2$ 
    \pause
    \begin{talign}
      \int &x^5\sqrt{1+x^2}\, dx 
      = \frac{1}{2} \int x^4\sqrt{u}\, du 
      = \mpause[1]{\frac{1}{2} \int (u-1)^2\sqrt{u}\, du }\\
      \mpause{&= \frac{1}{2} \int (u^2 -2u +1)\sqrt{u}\, du }
      \mpause{= \frac{1}{2} \int \big( u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + u^{\frac{1}{2}} \big)\, du } \\
      \mpause{&= \frac{1}{2} \big( \frac{2}{7}u^{\frac{7}{2}} - 2\cdot \frac{2}{5}u^{\frac{5}{2}} + \frac{2}{3}u^{\frac{3}{2}} \big) + C } \\
      \mpause{&= \frac{1}{7}(1+x^2)^{\frac{7}{2}} - \frac{2}{5}(1+x^2)^{\frac{5}{2}} + \frac{1}{3}(1+x^2)^{\frac{3}{2}}  + C }
    \end{talign}
  \end{exampleblock}
\end{frame}