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Integration Help

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the antiderivative:

    [tex]\int\frac{2x^3-5x^2+5x-12}{(x-1)^2(x^2+4)}[/tex]




    2. Relevant equations



    3. The attempt at a solution
    Using Integration by Partial Fractions:

    [tex]\int\frac{2x^3-5x^2+5x-12}{(x-1)^2(x^2+4)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{Cx+D}{(x^2+4)}[/tex]

    2x3-5x2+5x-12 = A(x-1)(x2+4)+B(x2+4)+(Cx+D)(x-1)2

    Multiplying out:

    = -4A+4Ax-Ax2+Ax3+Bx2+4B+Cx+D-2Cx2-2Dx-Cx3+Dx2

    Collecting like terms:

    2x3-5x2+5x-12 = x3(A-C)+x2(B-2C+D-A)+x(4A+C-2D) + (D+4B)

    Equating corresponding coefficients gives

    A-C = 2
    B-2C+D-A = -5
    4A+C-2D =5
    D+4B-4A=-12

    in the matrix form:

    1, 0, -1, 0, 2
    -1, 1, -2, 1, -5
    4, 0, 1, -2, 5
    -4, 4, 0, 1, -12

    I used matlab to get the reduced-row echelon form:

    rref=

    1, 0, 0, 0, 27
    0, 1, 0, 0, 8
    0, 0, 1, 0, 25
    0, 0, 0, 1, 64

    Therefore, A=27, B=8, C= 25 and D=64 (?)

    [tex]\int \frac{27}{(x-1)}+ \int \frac{8}{(x-1)^2}+ \int \frac{25x+64}{(x^2+4)}[/tex]

    = 27 Log(-1 + x)-8/(x-1)+32arctan(x/2)+25/2log(4+x2)

    I'm not sure if this is the correct answer to this problem, because I tried solving it using mathematica and I got:

    [tex]\frac{2}{(x-1)} + log(x-1) + (1/2) log (4+x^2)[/tex]

    I really appreciate it if someone could show me my mistakes. Thanks!
     
    Last edited: Mar 14, 2009
  2. jcsd
  3. Mar 15, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    [tex]2x^3-5x^2+5x-12 = A(x-1)(x^2+4)+B(x^2+4)+(Cx+D)(x-1)^2[/tex]

    is true for all values of x. Put x=1
    you'd get -10=5B =>B=-2. So I think you did something wrong in multiplying out.
     
  4. Mar 15, 2009 #3

    Mark44

    Staff: Mentor

    You got some of your coefficients wrong when you expanded the right side of your equation. After collecting like terms I got these coefficients:
    x^3: A + C (you have A - C)
    x^2: -A + B - 2C + D (same)
    x: 4A + C - 2D (same)
    1: -4A + 4B + D (you have 4B + D at first, but apparently added in the -4A term in later work)

    I didn't work this through, but I did find that B = -2.
     
  5. Mar 15, 2009 #4
    Thanks for your input guys! Changing "-1" to "1" made all the difference!

    I row-reduced the new matrix and got:

    A=1, B=-2, C=1 and D=0

    And eventually I got the right answer. Thanks very much for spotting my error.
     
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