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Math Help - Indefinite integral

  1. #1
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    Indefinite integral

    Can anyone help me with this integral?

    \int\frac{\1}{x^2-3}dx

    Thanks!
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  2. #2
    Super Member Quacky's Avatar
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    Re: Indefinite integral

    \int\frac{1}{x^2-a^2}~dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C

    You can generate a proof of this formula in reverse with the chain rule; I was presented with such a proof and then taught the standard form.
    Last edited by Quacky; January 15th 2012 at 06:41 AM.
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  3. #3
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    Re: Indefinite integral

    Thaaanks!
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  4. #4
    Member sbhatnagar's Avatar
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    Re: Indefinite integral

    You can use use partial fractions to prove that formula.

    \frac{1}{x^2-a^2}=\frac{1}{(x-a)(x+a)}

    \frac{1}{(x-a)(x+a)} \equiv \frac{A}{x-a}+\frac{B}{x+a}

    \implies 1 \equiv A(x+a)+B(x-a)

    Putting x= a, -a we obtain

    A=\frac{1}{2a} and B=-\frac{1}{2a}

    \frac{1}{(x-a)(x+a)} = \frac{1}{2a}\Big[ \frac{1}{x-a}-\frac{1}{x+a} \Big]

    \begin{align*} \int \frac{1}{x^2-a^2}dx &=\int \frac{1}{(x-a)(x+a)}dx=\frac{1}{2a}\int \left( \frac{1}{x-a}-\frac{1}{x+a} \right)dx \\ &= \frac{1}{2a} (\ln|{x-a}|-\ln{|x+a|})+C \\ &= \frac{1}{2a} \ln{\Big| \frac{x-a}{x+a} \Big|}+C \end{align*}
    Last edited by sbhatnagar; January 16th 2012 at 04:01 AM.
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  5. #5
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    Re: Indefinite integral

    It's perfect!! Thanks!
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  6. #6
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    Re: Indefinite integral

    I have a little problem: I did the same that you said with this integral (image), but when I checked it with Wolfram, there's a -1 which I don't know where come from.
    Attached Thumbnails Attached Thumbnails Indefinite integral-wolframalpha-20120115144355360.gif  
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: Indefinite integral

    What do you get if you do the long division?
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  8. #8
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    Re: Indefinite integral

    The same but without -x: 6ln|x+2|-2ln|x+a|
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  9. #9
    Member sbhatnagar's Avatar
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    Re: Indefinite integral

    Quote Originally Posted by gulamerasool View Post
    The same but without -x: 6ln|x+2|-2ln|x+a|
    \begin{align*} \frac{x-x^2}{(x+1)(x+2)} &= -\frac{(x^2+3x+2)-(4x+2)}{x^2+3x+2} \\ &= -1+\frac{4x+2}{(x+1)(x+2)} \\ &= -1+  \frac{4(x+1)-2}{(x+1)(x+2)} \\ &= -1+\frac{4}{x+2}-\frac{2}{(x+1)(x+2)} \\ &= -1+\frac{6}{x+2}-\frac{2}{x+2}\end{align*}

    Integrate each term separately and see what you will get.
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