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Math Help - looks simple Integral but it is fake

  1. #1
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    Red face looks simple Integral but it is fake

    Hi . Can anyone give me a way to solve this Integral :

    <br />
   \int \frac{x^2}{1+x^4} dx<br />

    Regards ...
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Try

    Try this...

    \int\frac{x^2}{x^4+1}dx=\int\frac{x^2-1+1}{x^4+1}dx=\int\frac{x^2-1}{x^4+1}dx+\int\frac{-1}{x^4+1}dx
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  3. #3
    MHF Contributor Unknown008's Avatar
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    EDIT:: Made a mistake. Re-checking everything...

    Sub x^2 = tan theta.
    x = sqrt(tan theta)
    dx = \frac12 (tan \theta)^{-\frac12}.sec^2\theta

    \int \frac{x^2}{1+x^4} dx = \int \frac{tan\theta}{1+tan^2 \theta} . \frac12 (tan \theta)^{-\frac12}.sec^2\theta d\theta

    \int \frac{tan\theta}{1+tan^2 \theta} . \frac12 (tan \theta)^{-\frac12}.sec^2\theta d\theta= \frac12 \int \frac{tan\theta}{sec^2\theta} . (tan \theta)^{-\frac12}.sec^2\theta d\theta

     \frac12 \int \frac{tan\theta}{sec^2\theta} . (tan \theta)^{-\frac12}.sec^2\theta d\theta= \frac12 \int tan\theta(tan \theta)^{-\frac12} d\theta  = \frac12 \int (tan\theta)^{\frac12} d\theta

     \frac12 \int (tan\theta)^{\frac12} d\theta =  \frac12 [\frac{2(tan\theta)^{\frac32}}{3sec^2\theta}]

    Not sure if that's possible...
    Last edited by Unknown008; June 12th 2010 at 01:03 PM.
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Unknown008 View Post
    You can use a trigonometric substitution...

    let x^4 = tan^2(\theta).

    \int \frac{x^2}{1+x^4} dx = \int \frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}} d\theta


    \int \frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}}d \theta= \int \frac{tan(\theta)}{\sqrt{sec^2(\theta)}}d\theta = \int \frac{tan(\theta)}{sec(\theta)} d\theta= \int sin(\theta)d\theta = -cos(\theta)

    -cos(\theta) = -\frac{1}{\sqrt{x^2+1}}
    How did you get a square root in the den?

    \int \frac{x^2}{1+x^4} dx = \int \frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}} d\theta

    What did you sub dx with?
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Oops, sorry, I made a big mistake... I messed things up... >.<

    Ok, I'll be looking for something else...
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  6. #6
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Unknown008 View Post
    Oops, sorry, I made a big mistake... I messed things up... >.<

    Ok, I'll be looking for something else...
    Lol it ok, I just kept trying and could not do it the way you tried. Thought I was going crazy lol. I tried subbing in a hyperbolic trig function and that did not work. I solved it using a power series but I'm not sure if that is what you are looking for?
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Try this...

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

    If the second integral has negative 1 in the numerator then

    x^2 - 1 - 1 DNE x^2
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  8. #8
    Ted
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    \int \frac{x^2}{1+x^4} \, dx

    =\int \left( \sum_{n=0}^{\infty} (-1)^n x^{4n+2} \right) \, dx

    =C+\sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+3}}{4n+3}
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