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Integral of improper fraction

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{x^{2}}{x^{2}-1}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I divide bottom into the top because the degree is the same and get...
    [tex]\int 1-\frac{1}{x^{2}+1}[/tex]

    My question is, from here, do I have to recognize the tan[tex]^{-1}[/tex]
    or is it also correct to put
    x-ln [tex]\left| x^{2} + 1\right|[/tex] + c ?
     
  2. jcsd
  3. Feb 11, 2009 #2

    statdad

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    Why would you use the logarithm when it isn't correct?
     
  4. Feb 11, 2009 #3

    HallsofIvy

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    Surely you know that you can't just "ignore" functions like that. The integral of 1/f(x), in general, has nothing to do with the integral of 1/x. Yes, you really have to be able to "recognize" the basic integrals.
     
  5. Feb 11, 2009 #4
    Never mind
     
    Last edited: Feb 11, 2009
  6. Feb 11, 2009 #5
    I never claimed to be overly bright. I asked because I was confused.
    Thanks for responding.
     
  7. Feb 11, 2009 #6

    HallsofIvy

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    Join the club! Most of us here are not overly bright and often confused.:tongue2:
     
  8. Feb 12, 2009 #7
    You only have to check the comment in my edit to realise that is true, my excuse is that I'd not long woken up and the government, ahem *cough*. :smile:
     
  9. Apr 3, 2009 #8
    can u help me with the integral of (2x^3)/(x-1)
     
  10. Apr 3, 2009 #9
    Whenever you have a troublesome integral, it always helps to start with trying u-substitution or integration by parts. Which technique do you think you should use here? Sometimes you just have to pick a method and try it out. If it fails, see if you can modify it so that it works. A further hint can be provided if you try these methods out and are still stuck.
     
  11. Apr 3, 2009 #10

    HallsofIvy

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    First, please do not "hijack" someone else's thread for a completely new question. Start your own thread using the "new post" button.

    Now, first divide [itex]2x^3+ 0x^2+ 0x+ 0[/itex] by x-1 to get a quadratic polynomial plus something of the form A/(x-1). Integrate that.
     
  12. Apr 3, 2009 #11
    I at first disregarded dividing by [itex]x-1[/tex], but it works out quickly (say with synthetic division or long division). If you have an idea, work it out! Letting [itex]u=x-1[/tex] works out fine, but just requires more algebra. Both methods give the same answer.

    Thank you for the forum advice HallsofIvy.
     
  13. Apr 3, 2009 #12
    Back to the original post. You have a few signs wrong (add everything back together and you don't get what you started with). After using long division to divide [itex]x^2-1[/tex] into [itex]x^2[/tex] (be sure that you can do this) you get:
    [tex]\int \frac{x^{2}}{x^{2}-1} \,dx = \int \left( 1 + \frac{1}{x^2-1} \right) \,dx = \int 1 \,dx + \int \frac{1}{(x-1)(x+1)} \,dx = x + C + \int \frac{1}{(x-1)(x+1)} \,dx[/tex]
    Now you expand the fraction on the far right using partial fractions. Then you will get integrals that can be evaluated using the natural log.
     
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