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Partial Fraction Decomp. Integral

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{4y^2-7y-12}{y(y+2)(y-3)}[/tex] with limits of integration [tex][1, 2][/tex]

    2. Relevant equations

    The answer given in the book is [tex]\frac{27}{5}ln(2)-\frac{9}{5}ln(3)[/tex]

    or [tex]\frac{9}{5}ln\frac{8}{3}[/tex]


    3. The attempt at a solution

    [tex]\int\frac{4y^2-7y-12}{y(y+2)(y-3)}[/tex]

    [tex]=\int\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}[/tex]

    I found:

    [tex]4y^2-7y-12=A(y+2)(y-3)+B(y)(y-3)+C(y)(y+2)[/tex]

    If y=-2, then [tex]B=1[/tex]

    I also found:

    [tex]4y^2-7y-12=(A+B+C)y^2+(2C-3B-A)y-6A[/tex]

    so [tex]6A=12[/tex], or [tex]A=2[/tex].

    Also, [tex](A+B+C)=4[/tex] so [tex](2+1+C)=4[/tex] and [tex]C=1[/tex]

    Therefore: [tex]=\int\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}[/tex]

    [tex]=\int\frac{2}{y}+\frac{1}{y+2}+\frac{1}{y-3}[/tex]

    [tex]=2ln|y|+ln|y+2|+ln|y-3|[/tex] with limits [tex][1, 2][/tex]

    [tex]=2(ln(2)-ln(1))+(ln(2+2)-ln(1+2)+(ln|2-3|-ln|1-3|)[/tex]

    [tex]=2ln(2)+(ln(4)-ln(3)+(ln(1)-ln(2))[/tex]

    [tex]=2ln(2)+ln(\frac{4}{3})-ln(2)[/tex]

    This is not equal to either answer provided in the book but I can't find an error. Also, how do you put the limits of integration into the function using Latex?
     
  2. jcsd
  3. Mar 21, 2010 #2

    Mark44

    Staff: Mentor

    Check your work. I get A = 2, B = 9/5, and C = 1/5.
     
  4. Mar 21, 2010 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    How do you get B=1? I get B=9/5. Also note you can get A (by putting y=0) and C (by putting y=3) the same way you got B.
     
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