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Homework Help: 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


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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 :(

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


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    Homework Helper

    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


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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:


    On the last integral rewritten as:


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