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Integration help needed

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

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

    i tried to get 'dx' into the integral but it didnt work out


    2. Relevant equations



    3. The attempt at a solution

    i have tried squaring the top and bottom, which gave {x[tex]^{4}[/tex] + 2x[tex]^{2}[/tex] +1}/{x^{2} + 2x +1}

    i dont know if that was a good idea, but i was stumped. I cant seem to get anything to cancel out
     
  2. jcsd
  3. Dec 8, 2007 #2
    hm, i'm not quite sure what your problem is

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

    is this correct?
     
  4. Dec 8, 2007 #3
    yea that's it.
     
  5. Dec 8, 2007 #4
    What integration technique are you learning at the moment? I'm just wondering so I can show you with that method.

    btw, it should be x^2 - 2x + 1 in the denominator.
     
  6. Dec 8, 2007 #5
    oh yeah you're right.
    well, so far we've learned the fundamental theorem of calculus and ordinary U substitutions. i havent gotten to the point of using special formulas for anything yet.
     
  7. Dec 8, 2007 #6
    Wow. This problem is giving me trouble, I feel ashamed!!!

    i have

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

    but it's not leading me anywhere, argh.
     
  8. Dec 8, 2007 #7

    malty

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    Gold Member

    Hey, join the club!!

    I got it down to [tex] 1+ \int{\frac{x+1}{x-1}}dx [/tex] is there anything that can be done here... ?




    ...wait I think i forgot to square it :(


    Yup:redface:
     
    Last edited: Dec 8, 2007
  9. Dec 8, 2007 #8

    Dick

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    Don't feel ashamed. Given what the OP said about the level of the course, this problem should never have been given before teaching partial fractions.
     
  10. Dec 8, 2007 #9
    Eh, it's doable with u-substitution. I don't want to just hand out the answer, but square and do polynomial division. Then you get a quadratic that's easy to integrate and something involving a u substitution.
     
  11. Dec 8, 2007 #10
    Thanks! I was feeling kinda crappy, lol. I'm actually in Calculus 2 so I should be able to do this, but it wasn't working out for me with simpler techniques.

    Well, I didn't think that. Uh!!! That should definitely work.
     
  12. Dec 8, 2007 #11

    Dick

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    Good point. That's the other way to do it.
     
  13. Dec 8, 2007 #12
    SOLVED! Now I''m happy :-] Thanks.
     
  14. Dec 8, 2007 #13
    when i used polynomial division i got big mess:

    (x^2)-2x+7+ [(12x-6)/((x^2)-2x+1)]

    so i have to use a U sub. now?
     
    Last edited: Dec 8, 2007
  15. Dec 8, 2007 #14
    from my polynomial division if i remember correctly, was different. let me re-do it real fast.
     
  16. Dec 8, 2007 #15
    Polynomial division, let's do this part first so our Integration will work out.

    [tex]\int\frac{x^4 +2x^2 +1}{x^2 -2x +1}dx=\int(x^2 +2x +5)+\frac{8x-4}{x^2 - 2x+1}dx[/tex]
     
    Last edited: Dec 8, 2007
  17. Dec 8, 2007 #16
    k isee that lol im an idiot. i havent done it in a while. i get it. but what about the U sub?
     
  18. Dec 8, 2007 #17
    k, I feel good about my Polynomial now.

    so the u-sub ...

    [tex]\int(x^2 +2x +5)+\frac{8x-4}{x^2 - 2x+1}dx[/tex]

    now splitting it into two Integrals

    [tex]\int(x^2 +2x +5)dx+\int\frac{8x-4}{x^2 -2x +1}dx[/tex]

    so now doing a u-substitution only for the second Integral

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

    [tex]u=x^2 -2x +1[/tex]

    now just take the derivative and you will need to mess around with the numerator a little more to get your du in your Integral.
     
    Last edited: Dec 8, 2007
  19. Dec 8, 2007 #18
    thanks man i really appreciate it. sorry for making you type all that up lol, but i got a calc final coming up
     
  20. Dec 8, 2007 #19
    No problem! You'll do fine :-] And the typing was no biggy, it was mainly copy/paste :-]
     
  21. Dec 9, 2007 #20
    Another possibility is to use the substitution:

    [tex]u=x-1[/tex]

    On the last integral rewritten as:

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

    I find this a bit easier. The different methods can be compared, all are equally valid.
     
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