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Thread: Intergration by parts

  1. June 12th 2009, 03:34 AM #1
    jamm89
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    Intergration by parts

    Can someone please check that i am doing this correctly
    \int \ 5xe^{4x}
    Let u = 5x
    v= e^{4x}/4
    du/dx = 5
    dv/dx = e^{4x}
    So 5x . e^{4x}/4 - \int e^{4x}/4 . 5 dx
    then \frac {\ 5xe^{4x}} {4} - \frac {5e^{4x}} {16}
    So factorised:
    \frac {\ 5e^{4x}} {4} x (x - 1/4) + c
    Cheers
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  2. June 12th 2009, 03:46 AM #2
    alexmahone
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    It is correct!

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