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Thread: 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 $\displaystyle
    dv = x^2 \sqrt {x^3 + 1} \Rightarrow \int {(dv} )dx = v
    $

    Then:

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

    $

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