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Thread: Indefinite Integral

  1. May 14th 2012, 07:18 AM #1
    Bashyboy
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    Indefinite Integral

    I attached a file containing the problem; I circled the answer, but it is not the correct answer. What did I do wrong?
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  2. May 14th 2012, 07:46 AM #2
    skeeter
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    Re: Indefinite Integral

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

    u = x-1

    x = u+1

    du = dx

    \int \frac{2(u+1)}{u^2} \, du = 2 \int \frac{1}{u} + \frac{1}{u^2} \, du

    integrate and back-substitute
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