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Math Help - Integration Question Help

  1. #1
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    Integration Question Help

    I need to integrate this:

    1/(x^4+4)

    I have that it factors to (x^2+2x+2)(x^2-2x+2). I did the partial fraction decomposition correctly (I think), and I got this:

    A=1/8
    B=1/4
    C=-1/8
    D=1/8

    I pulled 1/8 from the top of the integrations so I have to integrate:

    (1/8)* the integral (x+2)/(x^2+2x+2) - (1/8)*the integral (x+2)/(x^2-2x+2).

    That's where I'm stuck. Any help?
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  2. #2
    Behold, the power of SARDINES!
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    Try completing the square on the denominators. you should end up with some arctan in your solution.
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  3. #3
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    Yeah, I tried. I don't know what to do from completing the square.

    I get x^2+2x+1+1, which factors to (x+1)(x+1)+1.

    On the other, I get x^2-2x+1+1, which factors to (x-1)(x-1)+1.

    I don't know what to do with it.
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  4. #4
    Behold, the power of SARDINES!
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    \frac{1}{8}\int\frac{2-x}{x^2-2x+1}dx=\frac{1}{8}\int\frac{2-x}{(x-1)^2+1}dx

    then let u= x-1

    \frac{1}{8}\int\frac{2-(u+1)}{u^2+1}du=\frac{1}{8}\int\frac{1}{u^2+1}du-\frac{1}{8}\int\frac{u}{u^2+1}du
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  5. #5
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    Shouldn't it be (x+2) on top and not (2-x)?
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  6. #6
    Behold, the power of SARDINES!
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    My partial fractions could be wrong but I came up with

    \frac{1}{x^4+4}=\frac{1}{8}\cdot\frac{-x+2}{x^2-2x+2}+\frac{1}{8}\cdot\frac{x+2}{x^2+2x+2}
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  7. #7
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    Quote Originally Posted by thegame189 View Post
    I need to integrate this:

    1/(x^4+4)

    I have that it factors to (x^2+2x+2)(x^2-2x+2). I did the partial fraction decomposition correctly (I think), and I got this:

    A=1/8
    B=1/4
    C=-1/8
    D=1/8

    I pulled 1/8 from the top of the integrations so I have to integrate:

    (1/8)* the integral (x+2)/(x^2+2x+2) - (1/8)*the integral (x+2)/(x^2-2x+2).

    That's where I'm stuck. Any help?

    You have the same answer as me

    \frac{cx+d}{x^2-2x+2}=\frac{1}{8}\cdot\frac{-x+2}{x^2-2x+2}
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  8. #8
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    Does this seem right?

    (1/16)ln(x^2+2x+2) + (1/8)arctan(x+1) + (1/8)arctan(x-1) - (1/16)ln(x^2-2x+2) + C

    Something else said it should be -(1/8)arctan(1-x) as the third part instead of +(1/8)arctan(x-1). Which is it?.
    Last edited by thegame189; February 28th 2008 at 10:07 AM.
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  9. #9
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    ^^ Those two forms are equivalent. The - is distributed inside the arctan in the second (this is allowed since arctan is an odd function).

    edit: also, your solution is correct.
    Last edited by xifentoozlerix; February 28th 2008 at 11:23 AM.
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  10. #10
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    Yeah, that's what I was thinking. I just don't see why they'd write it with the negative. Oh well, thanks everyone. I was on the right track, I just got stuck after the square completing.
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