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Math Help - Integration by parts

  1. #1
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    Integration by parts

    Using integration by parts, would the following be correct?

    \int(2x-1)\ln x\,dx
    u = \ln x \Rightarrow du = \frac{1}{x}\,dx
    dv = (2x - 1)\,dx \Rightarrow v = \int 2x - 1\,dx = x^2 - x

    \int(2x - 1)\ln x\,dx = \ln x(x^2 - x) - \int{\frac{x^2 - x}{x}}\,dx

    = \ln x(x^2 - x) - \int{x - 1}\,dx
    = \ln x(x^2 - x) - \frac{x^2}{2} + x + C
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  2. #2
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    It's fine. Remember, you could always differentiate the anti-derivative that you arrive at and see if it returns the integrand.
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  3. #3
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    Ah, good point. Thanks.
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