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Homework Help: Integral of a fraction consisting of two quadratic equations

  1. Jul 5, 2008 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations
    I believe it's necessary to complete the square.

    3. The attempt at a solution
    Completing the square for





    [tex]u=x-\frac{1}{2}, du = dx, x=u+\frac{1}{2}[/tex]


    [tex]=4u + \ln|u^2+\frac{1}{2}|+c[/tex]

    Substituting back


    Would someone be so kind as to tell me if this is correct? For this question, I have 4 possible answers (and one that states "None of the above") and my answer doesn't match any of the other 3, so I'm wondering.

  2. jcsd
  3. Jul 5, 2008 #2
    I won't say if it's good or not since it seems like you're taking a quiz/exam or something. Differentiate to check your original Integral.
  4. Jul 5, 2008 #3


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    Science Advisor

    It seems to me you could simplify the problem by long division before completing the square.
  5. Jul 5, 2008 #4


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    Staff Emeritus
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    What happened to the factor of 4 in the denominator?
  6. Jul 6, 2008 #5
    Close...I'm busy with an assignment, but the whole thing only contributes about 5% to my year mark and is more for tutorial purposes than anything else since, where I'm studying, we don't have lectures at all. :smile:

    Thanks HallsofIvy, I'm going to give that a try today.

    Mmmm...Oh, wait, let me show you (perhaps I did something wrong here as well)

    [tex]4x^2-4x+3 = 0[/tex]

    Last edited: Jul 6, 2008
  7. Jul 6, 2008 #6
    \int \frac{4x^{2}-2x+2}{4x^{2}-4x+3} \mbox{d}x = \int \frac{1}{4x^{2}-4x+3}\ +\ \frac{2x-1}{4x^{2}-4x+3}\ \mbox{d}x=....
  8. Jul 7, 2008 #7


    PS. dirk_mec1, I think it should be

    [tex]\int 1 dx + \int\frac{2x-1}{4x^2-4x+3}dx[/tex]
  9. Jul 7, 2008 #8


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    [tex]\frac{1}{4x^{2}-4x+3}\ +\ \frac{2x-1}{4x^{2}-4x+3}= \frac{2x}{4x^2- 4x+ 3}[/tex]

    Did you mean
    \int \frac{4x^{2}-2x+2}{4x^{2}-4x+3} \mbox{d}x = \int 1\ +\ \frac{2x-1}{4x^{2}-4x+3}\ \mbox{d}x
  10. Jul 7, 2008 #9
    Yes sorry :shy:
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