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Math Help - Stumped by an Integral

  1. #1
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    Stumped by an Integral

    I am trying to evaluate the following:

    \int\frac{2+x^2}{1+x^2}dx

    So far I have:

    \int\frac{2}{1+x^2}dx+\int\frac{x^2}{1+x^2}dx

    =2tan^{-1}(x)+\int\frac{x^2}{1+x^2}dx

    Obviously I'm stuck on the second term. Any ideas?
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  2. #2
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    \int\frac{2+x^2}{1+x^2}dx=\frac{1}{2}\int\frac{1+u  }{\sqrt{u}(1+u)}du=\frac{1}{2}\int\frac{1}{\sqrt{u  }}du

    u=x^2



    nevermind,
    Last edited by adkinsjr; December 7th 2009 at 03:46 PM.
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  3. #3
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    \int\frac{2+x^2}{1+x^2}dx = \int\frac{1+x^2 + 1}{1+x^2}dx = \int 1 +\frac{1}{1+x^2} dx

    And \int\frac{1+x^2}{1+x^2}dx = \int 1dx
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  4. #4
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    Yeah, that last one was a typo. I edited it. I meant for 2-x^2 to be in the numberator. thanks
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  5. #5
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    Oops, I misunderstood your last post.
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