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Math Help - integrating

  1. #1
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    integrating

    how do you integrate

    4u / (1+u)^2 (1+u^2) du
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    how do you integrate

    4u / (1+u)^2 (1+u^2) du
    do you mean:

    [4u/(1 + u)^2] * (1 + u^2) ??
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  3. #3
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    oh no i meant

    4u /[ (1+u)^2 * (1+u^2)] du

    sorry didnt know it could have been interpreted like that o.o
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  4. #4
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    Hello, jeph!

    How do you integrate:
    . . . . . . . . 4u
    . . ∫ -------------------- du
    . . . (1 + u)(1 + u)

    You need Partial Fractions . . .

    . . . . . 4u . . . . . . . . . . A . . . . . .B . . . . .Cu + D
    -------------------- . = . ------- + ---------- + ---------
    (1 + u)(1 + u) . . . . 1 + u . . (1 + u) . . 1 + u

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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    how do you integrate

    4u / (1+u)^2 (1+u^2) du
    Here
    Attached Thumbnails Attached Thumbnails integrating-parfrac-int.gif  
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  6. #6
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    Re;

    Re:
    Attached Thumbnails Attached Thumbnails integrating-7.gif  
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  7. #7
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    Quote Originally Posted by qbkr21 View Post
    Re:
    Once you try to write the partial fractions, you have to respect this rule:
    Denominator Containing Repeated Linear Factors

    If a linear factor is repeated n times in the denominator, there will be n corresponding partial fractions with degree 1 to n.
    You have to have that because you have to have the same polynomial as a denominator. Your expresion, does not have the same degree as the initial denominator.
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  8. #8
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    Re:

    Re:
    Attached Thumbnails Attached Thumbnails integrating-13.gif  
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Re:
    yes, that's correct
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  10. #10
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    Re:

    Re:
    Attached Thumbnails Attached Thumbnails integrating-14.gif  
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Re:
    we can find C immediately by plugging in x = -1, but for the rest, it seems you would have to multiply out to get them
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  12. #12
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    Re:

    RE:

    Yea I found c to be equal to (-1/4)

    I will post my work as I progress along on this problem tonight!

    Thanks!
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  13. #13
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    Quote Originally Posted by Soroban View Post
    Hello, jeph!


    You need Partial Fractions . . .

    . . . . . 4u . . . . . . . . . . A . . . . . .B . . . . .Cu + D
    -------------------- . = . ------- + ---------- + ---------
    (1 + u)(1 + u) . . . . 1 + u . . (1 + u) . . 1 + u
    Before to apply partial fractions, I'd check first that (1+u)^2-(1+u^2)=2u, which frees to us of the previous mechanism.

    Regards
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