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Math Help - Integration by partial fractions.

  1. #1
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    Integration by partial fractions.



    Ive been able to complete a couple of problems. I was hoping someone could help me with this one. Thanks.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by xcelxp View Post


    Ive been able to complete a couple of problems. I was hoping someone could help me with this one. Thanks.
    First, I would use polynomial division to simplify this fraction:

    4x^2 - 4 | 3x^3 - 0x^2 - 3x + 4
    = 3/4*x + 4/(4x^2 - 4)
    = 3/4*x + 1/(x^2 - 1)

    So now we have:

    INT (3/4*x + 1/(x^2 - 1)) dx
    = INT 3/4*x dx + INT 1/(x^2 - 1) dx
    = 3/8*x^2 + INT 1/(x^2 - 1) dx

    Use Partial fractions on the remaining integration:
    INT 1/(x^2 - 1) dx

    Factor the denominator:
    x^2 - 1 = (x - 1)(x + 1)

    Now set up your seperate fractions:

    1/(x^2 - 1) = A/(x - 1) + B/(x + 1)

    Multiply out the denominator (x^2 - 1) to both sides:

    1 = A(x + 1) + B(x - 1) = Ax + A + Bx - B

    Set terms equal on both sides (x's with x's and constants with constants)

    (A + B)x = 0x --> A + B = 0 --> A = -B
    (A - B) = 1 --> -B - B = 1 --> B = -1/2, A = 1/2

    So the integration becomes:

    INT 1/2*1/(x - 1) - 1/2*1/(x + 1) dx = 1/2ln|x - 1| - 1/2ln|x + 1|


    The final answer is:
    3/8*x^2 + 1/2ln|x - 1| - 1/2ln|x + 1|

    Edited: My original solution had 2 as the coefficients of the ln functions when they should be 1/2.
    Last edited by ecMathGeek; April 20th 2007 at 02:23 PM.
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by ecMathGeek View Post
    3/8*x^2 + 2ln|x - 1| - 2ln|x + 1|
    You forgot the constant. Also I think that the solution would be

    (3/8)x + (1/2)ln|x - 1| - (1/2)ln|x + 1| + c.

    Greetings
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Krizalid View Post
    You forgot the constant. Also I think that the solution would be

    (3/8)x + (1/2)ln|x - 1| - (1/2)ln|x + 1| + c.

    Greetings
    Thank you for pointing out the constant, but are you sure about the (1/2) coefficients? I can go back through my work and see if I missed something, but I'm fairly confident they are 2's and not 1/2's.

    You're right. I had a typo in my work that I carried through by accident. I might be off today or something. This isn't my first typo.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Thank you for pointing out the constant, but are you sure about the (1/2) coefficients? I can go back through my work and see if I missed something, but I'm fairly confident they are 2's and not 1/2's.
    i agree with krizalid, i think the coefficients are (1/2)'s. let me double check

    Quote Originally Posted by ecMathGeek View Post

    Now set up your seperate fractions:

    4/(x^2 - 1) = A/(x - 1) + B/(x + 1)

    Multiply out the denominator (x^2 - 1) to both sides:

    4 = A(x + 1) + B(x - 1) = Ax + A + Bx - B
    i believe here is where you made an error. you have the numerator as 4 here, when you should have it as 1 according to your method

    Quote Originally Posted by ecMathGeek View Post
    I might be off today or something. This isn't my first typo.
    I have days like that too--and worst days. not only do i make typos on such days, but i misread the question in the first place.

    so i try to solve the wrong question AND i do it incorrectly...luckily you guys are always there to back me up
    Last edited by ThePerfectHacker; April 24th 2007 at 06:42 PM.
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  6. #6
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    Hello, xcelxp!

    I was taught a different procedure for Partial Fractions.
    . . Try it . . . I think you'll like it.

    . . . . . . . .3x - 3x + 4 . . . . . . . . 1
    We have: .-------------- .= .x + -------
    . . . . . . . . .4x - 4 . . . . . . . . . .x - 1


    Now we decompose that fraction:

    . . . . . . 1 . . . . . . . . .A . . . . B
    . . ---------------- .= .------ + ------
    . . (x - 1)(x + 1) . . . x - 1 . .x + 1


    Clear denominators: .1 .= .A(x + 1) + B(x - 1)

    .Let x = 1: .1 .= .A(2) + B(0) . . . A =

    Let x = -1: .1 .= .A(0) + B(-2) . . B = -

    . . . . . . . . . . . . . . . . . . . . . . . . . . . .
    Hence, we must integrate: .x + ------ - ------
    . . . . . . . . . . . . . . - . . . . . . . . .x - 1 - x + 1

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