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Math Help - integrate ln function

  1. #1
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    integrate ln function

    integrate ln(x^2-1)
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  2. #2
    Eater of Worlds
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    Use the rules of exponents to rewrite as

    ln(x^{2}-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)
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  3. #3
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    Quote Originally Posted by galactus View Post
    Use the rules of exponents to rewrite as

    ln(x^{2}-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)
    But care needs to be taken with domain issues since the original function is defined for x > 1 or x < -1 whereas the re-write has problems when x < -1 ....
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  4. #4
    Eater of Worlds
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    Yep. I was being lazy and careless.
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  5. #5
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    so how should I go about solving this problem?
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  6. #6
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    Do exactly what galactus said but be careful like mr fantastic said!

    ln(x^2- 1) is an even function so you really only need to integrate for x> 1, where galactus' method has no problem, and then use the same integral for x< -1.
    Last edited by mr fantastic; February 8th 2009 at 06:48 PM. Reason: Fixed the latex tag
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  7. #7
    Member javax's Avatar
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    Or you can go with integration by parts

    \int{\ln(x^2-1)}

    sub. \ln(x^2-1)=u \Rightarrow \frac{2x dx}{x^2-1}=du; dv = dx \Rightarrow v = x;

    so we have

    x\ln(x^2-1)-2\int{\frac{x^2}{x^2-1}dx}
    = x\ln(x^2-1)-2\int{\frac{x^2-{\color{red}1}+{\color{red} 1}}{x^2-1}dx}
    =x\ln(x^2-1)-2\int{\left (1+\frac{1}{x^2-1}\right )dx} =...

    correct me anyone if I did anything wrong.
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