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Math Help - Integration of quadratic denominaters error

  1. #1
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    Integration of quadratic denominaters error

    Hello

    im stuck on the following integral

    \displaystyle\int^9_5 \dfrac {1}{10x-9-x^2} \, dx

    ill go through how iv done it which seems correct but then when i get to inputting the limits it produces math error on my calculater so i must have done something wrong.

    \displaystyle\int^9_5 \dfrac {1}{-(x^2-10x) -9} \, dx

    \displaystyle\int^9_5 \dfrac {1}{-(x^2-10x +25) +25 -9} \, dx

    \displaystyle\int^9_5 \dfrac {1}{-(x-5) +16} \, dx

    \displaystyle\int^9_5 \dfrac {1}{4^2 - (x-5)^2} \, dx

    so since

    \displaystyle\int \dfrac {dx}{a^2 - x^2} = \dfrac{1}{a}arctanh \dfrac {x}{a} +C we get \dfrac{1}{4}arctanh \dfrac {x-5}{4} +C

    but when x = 9 acrtanh 1 which produces and error and when x = 5 acrtanh 0 which is 0

    any help on this would be most appreciated
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  2. #2
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    Quote Originally Posted by jordanrs View Post
    Hello

    im stuck on the following integral

    \displaystyle\int^9_5 \dfrac {1}{10x-9-x^2} \, dx

    ill go through how iv done it which seems correct but then when i get to inputting the limits it produces math error on my calculater so i must have done something wrong.

    \displaystyle\int^9_5 \dfrac {1}{-(x^2-10x) -9} \, dx

    \displaystyle\int^9_5 \dfrac {1}{-(x^2-10x +25) +25 -9} \, dx

    \displaystyle\int^9_5 \dfrac {1}{-(x-5) +16} \, dx

    \displaystyle\int^9_5 \dfrac {1}{4^2 - (x-5)^2} \, dx

    so since

    \displaystyle\int \dfrac {dx}{a^2 - x^2} = \dfrac{1}{a}arctanh \dfrac {x}{a} +C we get \dfrac{1}{4}arctanh \dfrac {x-5}{4} +C

    but when x = 9 acrtanh 1 which produces and error and when x = 5 acrtanh 0 which is 0

    any help on this would be most appreciated
    There is nothing wrong with one of the limits giving a result of 0.

    For the one where arctanh is not defined, I suggest using:

     \displaystyle \lim_{n \to \infty} \dfrac{1}{4}\text{arctanh} \dfrac {x-5}{4} +C \bigg|_5^n
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  3. #3
    Member Nacho's Avatar
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    Maybe this an improper integral, which diverge
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  4. #4
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    yeah good point about it being undefined / going of to infinity didnt think about that. cheers for the help
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