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\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}