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      <text>13.4.1.3 By parts multiple times</text>
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      <text><![CDATA[<p>Sometimes it is necessary to apply the formula more than once, as the next Example shows.</p>
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<h4 class="HELM_example">Example</h4>
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<p>Find \({\displaystyle\int 2x^2 e^{-x} \, \mathrm{d}x}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>We let \(f=2x^2\) and \(g=e^{-x}\). Then \(\dfrac{\mathrm{d} f}{\mathrm{d} x}=4x\) and \({\displaystyle\int g \, \mathrm{d}x} = -e^{-x}\)</p>
<p>Using the formula for integration by parts we find</p>
\({\displaystyle\int 2x^2 e^{-x} \, \mathrm{d}x = 2x^2(-e^{-x})-\displaystyle\int 4x(-e^{-x}) \, \mathrm{d}x = -2x^2e^{-x}+\displaystyle\int4xe^{-x} \, \mathrm{d}x}\)

<p>We now need to find \({\displaystyle\int4xe^{-x} \, \mathrm{d}x}\) using integration by parts again. We get</p>

\begin{eqnarray*} \int 4xe^{-x}\ \mathrm{d} x&amp;=&amp;4x(-e^{-x})-\int4(-e^{-x})\ \mathrm{d} x\\ &amp;=&amp;-4xe^{-x}+\int 4e^{-x}\ \mathrm{d} x\\ &amp;=&amp;-4xe^{-x}-4e^{-x}\end{eqnarray*}
<p>Altogether we have</p>

\({\displaystyle\int 2x^2e^{-x} \, \mathrm{d}x} = -2x^2e^{-x}-4x e^{-x}-4 e^{-x}+c = -2e^{-x}(x^2+2x+2)+c\)]]></text>
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