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Math Help - Help with an integral

  1. #1
    Member
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    May 2008
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    Help with an integral

    Hey I'm having a little trouble with this integral
    <br />
\int \frac{1+x^2}{1+x^4}dx

    I know the answer just need a little nudge in the right direction with it

    many thanks

    Simon

    EDIT dont worry I got it, stupid me forgetting
    <br />
(1+x^4)=(1-\sqrt 2 x+x^2)(1+\sqrt 2 x+x^2)
    Last edited by thelostchild; May 29th 2008 at 03:57 PM.
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  2. #2
    Super Member
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    Quote Originally Posted by thelostchild View Post
    Evaluate the following integral, without using partial fractions

    \int \frac{1+x^2}{1+x^4}dx

    first divide by x^2

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

    Add zero to the denominator by adding and taking away 2

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

    Compete "the square"

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

    Now use the subistution u = x - x^{-1} \ \ \ \Rightarrow \ \ \ du = (1+x^{-2})dx

    that makes the integral. \int \frac{1}{u^2 + 2}du

    use another subistution u = \sqrt{2} \tan \theta

    (skipping some steps) ....

    \int \frac{1}{u^2 + 2}du = \frac{1}{ \sqrt{2}}\arctan{\frac{u}{ \sqrt{2}}} + C


    \therefore \int \frac{1+x^2}{1+x^4}dx = \frac{1}{ \sqrt{2}}\arctan \left( \frac{ x - x^{-1}}{ \sqrt{2}} \right) + C

    Bobak
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