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Thread: Intergration by parts

  1. #1
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    Intergration by parts

    Can someone please check that i am doing this correctly
    $\displaystyle \int \ 5xe^{4x}$
    $\displaystyle Let u = 5x $
    $\displaystyle v= e^{4x}/4$
    $\displaystyle du/dx = 5$
    $\displaystyle dv/dx = e^{4x}$
    So $\displaystyle 5x . e^{4x}/4 $ - $\displaystyle \int e^{4x}/4 . 5 dx$
    then $\displaystyle \frac {\ 5xe^{4x}} {4} - \frac {5e^{4x}} {16}$
    So factorised:
    $\displaystyle \frac {\ 5e^{4x}} {4} $ x $\displaystyle (x - 1/4) + c $
    Cheers
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  2. #2
    MHF Contributor alexmahone's Avatar
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