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Math Help - Another Inegration by Parts

  1. #1
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    Another Inegration by Parts

    Pauls Online Notes : Calculus II - Integration by Parts

    Click the link and scroll down.
    Look at Example 7. In this example why does v = 2/9 ... I do not see were the 2/9 comes from. I also do not see where the -4/45 comes from.
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  2. #2
    Member Nacho's Avatar
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    If <br />
dv = x^2 \sqrt {x^3  + 1}  \Rightarrow \int {(dv} )dx = v<br />

    Then:

    <br />
\begin{gathered}<br />
  \int {x^2 \sqrt {x^3  + 1} } dx \hfill \\<br />
  u^2  = x^3  + 1 \Rightarrow 2udu = 3x^2 dx \Leftrightarrow dx = \frac{{2u}}<br />
{{3x^2 }}du \hfill \\<br />
  \int {\frac{2}<br />
{3}\sqrt u  \cdot udu = \frac{2}<br />
{3}\int {u^2 du}  = \frac{2}<br />
{3} \cdot \frac{{u^3 }}<br />
{3}}  = \frac{2}<br />
{9} \cdot \left( {u^2 } \right)^{3/2}  = \frac{2}<br />
{9} \cdot \left( {x^3  + 1} \right)^{3/2}  = v \hfill \\ <br />
\end{gathered} <br /> <br />

    I donīt understand why use integration by parts, is more hard than use the subtitution u^2=x^3+1
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