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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}