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Integration by partial fractions

  1. Jan 22, 2006 #1
    I started this section off quite well and I did very well on the problems where there are only linear factors but when I got to the problems with quadratic factors, I began getting wrong answers. I though that perhaps I would receive some advice or my error/mistake could be corrected if possible.

    The problem is:

    It comes with a small help thing that says: [tex](x-3)(x^2-6x+13) =

    Ok so the first thing I did was use long division and I got [tex]4x-7[/tex] as the answer of the long division. Next I proceeded to do:
    [tex]4x^4-43x^3+200x^2-442x+383 = A(x^2-6x+13) + (Cx+D)(x-3)[/tex].
    Next, I plugged in 3 for the value of x so that I could solve for A and I got its value to be 5. Next, I plugged in 0 for the value of x so that I could get A and D alone and solve for D. I got its value to be -106, which is horribly wrong for some reason, I think. Next I plugged in 1 for the value of x and solved for C nd its value came out to be 75, which is also horribly wrong probably..I made a mistake somewhere in there I think...after that, I set up the new integral:
    I solved that and got my final answer to be:

    This answer is wrong and the correct answer is:

    If I am making some kind of mistake, then I must be inherently making it because all my other answers to problems with quadratic factors is coming out wrong...please help me.
  2. jcsd
  3. Jan 22, 2006 #2
    You should have a remainder after you divide:
    [tex]4 x - 7 + \frac{13 x^2 - 69 x + 10}{x^3 - 9 x + 31 - 39}[/tex]
    That's an awfully tedious problem. :yuck:
  4. Jan 22, 2006 #3
    Yes, I do have a remainder after I divide. After the long division, I have:
    [tex]4 x - 7 + \frac{13 x^2 - 69 x + 110}{x^3 - 9 x^2 + 31x - 39}[/tex]

    What do I do with it? I was just looking over that because I noticed something about it in the book, but I can't quite grasp what the remainder would be used for..
    Last edited: Jan 22, 2006
  5. Jan 22, 2006 #4
    You'll want to split the remainder into partial fractions and then you can integrate everything.
  6. Jan 22, 2006 #5
    Ok, I just tried that and I finally got the right answer...I guess that's what my error was. After the long division, I need to use the remainder..heh..
    Last edited: Jan 22, 2006
  7. Jan 22, 2006 #6
    Upon further analysis, I realized that the answer is still wrong, this is why:

    after the division and the [tex]a^2 + x^2[/tex] trig substitution, the integral is:
    \int\frac{16\tan\theta+24-15}{4\sec^2\theta}2\sec^2\theta d\theta[/tex]

    The 8 in front of the natural log is supposed to be a four, anyone know why I have an 8?...
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