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Math Help - Integral: have madness, need method

  1. #1
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    Integral: have madness, need method

    <br />
\frac{x - 3}{x^2 + 1}<br />

    This looks like a tan^-1x type answer but the actual answer is different and I am not sure of the process to get there.
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  2. #2
    Member Mentia's Avatar
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    okay so separate it out:

     \frac{ x-3}{x^{2}+1 } =  \frac{x }{ x^2+1} - \frac{ 3}{x^2+1 }

    then

    <br />
 \int_{ }^{ }  \frac{ x-3}{x^{2}+1 }dx = \int_{ }^{ }   \frac{x }{ x^2+1}dx -  \int_{ }^{ } \frac{ 3}{x^2+1 }dx

    The part with the 3 is just arctan. Use a substitution on the other part:
    u = x^2 + 1 --> du = 2xdx ---> dx = du/(2x), and so the integral should give a natural log.
    Last edited by mr fantastic; March 8th 2009 at 05:29 AM.
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  3. #3
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    Quote Originally Posted by Mentia View Post
    Whoops, that funny part is:

    <br />
 \int_{ }^{ }  \frac{ x-3}{x^{2}+1 }dx = \int_{ }^{ }   \frac{x }{ x^2+1}dx -  \int_{ }^{ } \frac{ 3}{x^2+1 }dx
    haha thank goodness I didn't have to try to sort out the first post lol. This is so obvious no that I see it

    thank you so much for taking the time
    Last edited by mr fantastic; March 8th 2009 at 05:28 AM. Reason: Added oness ....
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