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Math Help - [SOLVED] Integrating log function

  1. #1
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    [SOLVED] Integrating log function



    my u = (x^2+1), du = 2x

    4 \int (2x+1)/(x^2+1)

    I would like to turn that into du/u but there's a + 1 in the numerator, which isn't part of du's value. What am I supposed to do?
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  2. #2
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    Quote Originally Posted by Archduke01 View Post


    my u = (x^2+1), du = 2x

    4 \int (2x+1)/(x^2+1)

    I would like to turn that into du/u but there's a + 1 in the numerator, which isn't part of du's value. What am I supposed to do?
    Try breaking up the fraction into 2 fractions

    \frac{8x+4}{x^2+1}=\frac{8x}{x^2+4}+\frac{4}{x^2+4  }
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  3. #3
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    Thank you!
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  4. #4
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    \int{\frac{2x+1}{x^2+1}}dx=\int{\frac{2x}{x^2+1}dx  +\int{\frac{1}{x^2+1}}}dx

    \int{\frac{2x}{x^2+1}}dx=\int{\frac{du}{u}}
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