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Math Help - Partial Fractions HELP PLEASE!!!

  1. #1
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    Partial Fractions HELP PLEASE!!!

    I have a problem solving this partial fraction... Please somebody help me out...


    Integrate the following partial fraction:

    2X + 2
    __________________ dx
    (x^2 + 1 ) (x + 1)^3
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  2. #2
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    Perform a partial fraction expansion and integrate each term individually. Where are you stuck?
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  3. #3
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    How about this?

    2x+2 dx
    (x^2+1)(x+1)^3

    Couldn't you do this:

    .......x+1
    2-------------------dx
    (x^2+1)(X+1)^3

    .........1
    2-------------------dx
    (x^2+1)(X+1)^2

    then 2 * (x^2+1)^-1 * (x+1)^-2 dx

    Then I assume multiply it out an integrate normally.
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  4. #4
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    2X + 2
    __________________ dx
    (x^2 + 1 ) (x + 1)^3

    Rewriting that in one line only,
    (2x+2)/[(x^2 +1)(x+1)^3] dx

    Simplifying it because there is a factor (x+1) that is common to both numerator and denominator,
    = 2(x+1)/[(x^2 +1)(x+1)^3] dx
    = 2/[(x^2 +1)(x+1)^2] dx

    You want to decompose 2/[(x^2 +1)(x+1)^2] into partial fractions.

    Okay.
    I hope you know the process so that I don't have to explain it. (If not, then ask me.)

    2/[(x^2 +1)(x+1)^2] = (Ax +B)/(x^2 +1) +C/(x+1) +D/(x+1)^2 ....(1)
    Multiply both sides by [(x^2 +1)(x+1)^2],
    2 = (Ax +B)(x+1)^2 +C(x^2 +1)(x+1) +D(x^2 +1) ....(2)

    When x=(-1), plug that into (2),
    2 = (A(-1) +B)(-1+1)^2 +C(x^2 +1)(-1+1) +D((-1)^2 +1)
    2 = 0 +0 +D(2)
    So,
    D = 2/2 = 1 ....***

    When x=0, plug that into (2), ...and plug also D=1,
    2 = (A(0) +B)(0+1)^2 +C(0^2 +1)(0+1) +1(0^2 +1)
    2 = (B)(1) +C(1)(1) +1(1)
    2 = B +C +1
    2 -1 = B +C
    1 = B +C ....(3)

    When x=1, plug that into (2), ...and plug also D=1,
    2 = (A(1) +B)(1+1)^2 +C(1^2 +1)(1+1) +1(1^2 +1)
    2 = (A+B)(2)^2 +C(2)(2) +1(2)
    2 = (A+B)(4) +C(4) +2
    2 -2 = 4(A+B) +4C
    0 = 4(A+B) +4C
    0 = (A+B) +C
    0 = A +B +C ...(4)

    Eq.(4) minus Eq.(3),
    -1 = A ....***

    When x=2, plug that into (2), ...and plug also D=1 and A = -1,
    2 = (-1(2) +B)(2+1)^2 +C(2^2 +1)(2+1) +1(2^2 +1)
    2 = (-2 +B)(3)^2 +C(5)(3) +1(5)
    2 = (-2 +B)(9) +15C +5
    2 = -18 +9B +15C +5
    2 +18 -5 = 9B +15C
    15 = 9B +15C
    Divide both sides by 3,
    5 = 3B +5C ....(5)

    From (3), C = 1-B, plug that into (5),
    5 = 3B +5(1-B)
    5 = 3B +5 -5B
    5 -5 = -2B
    B = 0 ....***

    Susbtitute that in (3),
    C = 1 -0 = 1 ...***

    So we found
    A = -1
    B = 0
    C = 1
    D = 1

    Hence,
    2/[(x^2 +1)(x+1)^2] = (Ax +B)/(x^2 +1) +C/(x+1) +D/(x+1)^2 ....(1)
    2/[(x^2 +1)(x+1)^2] = (-1x +0)/(x^2 +1) +1/(x+1) +1/(x+1)^2
    2/[(x^2 +1)(x+1)^2] = -x/(x^2 +1) +1/(x+1) +1/(x+1)^2 .....******

    There, it is decomposed.

    ------------------
    So,
    INT.[(2x +2)/[(x^2 +1)(x+1)^3]]dx
    = INT.[2 /[x^2 +1)(x+1)^2]]dx
    = INT.[-x/(x^2 +1) +1/(x+1) +1/(x+1)^2]dx
    = -(1/2)ln(x^2 +1) +ln(x+1) -1/(x+1) +C ....answer.
    Last edited by ticbol; April 20th 2005 at 11:13 AM.
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  5. #5
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    Thank you guys for the help. This is one of the partial fraction problems where one little mistake will cost you millions of headaches.
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