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Math Help - Tough Integral (Trig Sub?)

  1. #1
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    Tough Integral (Trig Sub?)

    Little help with this one? I can't quite see where its going.

    ∫2dx/(x^2-2x)

    Thanks ya'll
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  2. #2
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    Quote Originally Posted by Audioslavery View Post
    Little help with this one? I can't quite see where its going.

    ∫2dx/(x^2-2x)

    Thanks ya'll
    Try completing the square for the denominator then use this standard integral:

     \int \frac{du}{u^2-a^2} = \frac{1}{2a} ln \left|\frac{u-a}{u+a} \right| + C
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  3. #3
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    Quote Originally Posted by Audioslavery View Post
    Little help with this one? I can't quite see where its going.

    ∫2dx/(x^2-2x)

    Thanks ya'll
    Partial fractions would allow you to integrate without using tables.
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    Quote Originally Posted by dwsmith View Post
    Partial fractions would allow you to integrate without using tables.
    Ah thank you, I used the partial fractions method (although I still lack confidence using that method, algebra was a big fail for me).

    I ended up getting -ln|x| + ln|x-2|, I still need to evaluate this from 3 to ∞, I'm a little shaky regarding ∞ and natural log.

    so it ends up being [-ln|∞| + ln|∞-2|] - [ln|3| + ln|3-2|]

    the final answer is supposedly LN|3| according to my book, am I on the right track?
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  5. #5
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    Quote Originally Posted by Audioslavery View Post
    Ah thank you, I used the partial fractions method (although I still lack confidence using that method, algebra was a big fail for me).

    I ended up getting -ln|x| + ln|x-2|, I still need to evaluate this from 3 to ∞, I'm a little shaky regarding ∞ and natural log.

    so it ends up being [-ln|∞| + ln|∞-2|] - [ln|3| + ln|3-2|]
    Before you evaluate, combine your ln functions and then use L'Hopital's Rule.
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