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Math Help - (Integral) making the numerator = 1

  1. #1
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    (Integral) making the numerator = 1

    \int \frac {5x-6}{x^2-4x+29} dx

    I complete the square and get; \int \frac {5x-6}{(x-2)^2 + 25} dx

    If I can make the numerator equal to 1, I can use the following rule; \int \frac {1}{x^2+a^2} dx = \frac {1}{a} arctan \frac {x}{a} + C

    But I'm confused as to how to do that.
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    \int \frac {5x-6}{x^2-4x+29} dx

    I complete the square and get; \int \frac {5x-6}{(x-2)^2 + 25} dx

    If I can make the numerator equal to 1, I can use the following rule; \int \frac {1}{x^2+a^2} dx = \frac {1}{a} arctan \frac {x}{a} + C

    But I'm confused as to how to do that.
    First write the numerator, 5x- 6 as 5(x- 2)+ 4 so the integral is
    \int \frac{5(x-2)+ 4}{(x-2)^2+ 25} dx.

    Now separate it into two integrals:
    5\int \frac{x-2}{(x-2)^2+ 25}dx+ 4 \int\frac{1}{(x-2)^2+ 25} dx.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    First write the numerator, 5x- 6 as 5(x- 2)+ 4
    Is there some process you used to come up with that rewrite or is it something I'll simply have to have a feel for? I understand your following steps though, thanks a lot.
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