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Thread: Integration by Parts again

  1. #1
    Junior Member symstar's Avatar
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    [solved]Integration by Parts again

    Not quite sure what to pick for u and dv.
    \int_{1}^{2}(\ln{x})^2dx

    I considered u=\ln{x} and dv=x^{2}dx i.e considering it as a function in a function. Can I do this; Should I do this?

    Edit: Over-complicating the situation as usual. I found my problem and solved it.
    Last edited by symstar; Sep 2nd 2008 at 09:17 PM.
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