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Integration Problem

  1. Nov 28, 2005 #1
    I have a question in my book to evaluate
    [tex]\int_{2}^{3}_\frac{x^2-2x+3}{(x-1)^2}dx[/tex]
    I have completed the square on the top line of the fraction so that
    [tex]\frac{x^2-2x+3}{(x-1)^2}[/tex] = [tex]\frac{(x-1)^2+2}{(x-1)^2}[/tex]
    I have simplified so I end up with [tex]\int_{2}^{3}1+\frac{2}{(x-1)^2}dx[/tex] and now I don't know how to integrate this function. I have never encountered an integral with a number other than one on top of the fraction.
     
    Last edited: Nov 28, 2005
  2. jcsd
  3. Nov 28, 2005 #2

    VietDao29

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    You can use u-substitution in this case.
    So let u = x - 1, so du = dx.
    The integral becomes:
    [tex]\int \left( 1 + \frac{2}{(x - 1) ^ 2} \right) dx = \int dx + \int \frac{2}{(x - 1) ^ 2} dx = x + \int \frac{2}{u ^ 2} du[/tex], now after integrating: [tex]\int \frac{2}{u ^ 2} du[/tex], just change u back to x. After finding the anti-derivative of the function, just apply the limits to it, and you are done. Can you go from here?
     
    Last edited: Nov 28, 2005
  4. Nov 28, 2005 #3
    Thanks for the help but the question is from the chapter before the substitution method so I don't think substitution is to be used. Is there another way to do it?
     
  5. Nov 28, 2005 #4

    TD

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    Perhaps you have used subtitution implicitly. As you said, we can simplify it to:

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

    Now rewrite to see:

    [tex]\int {\frac{2}{{\left( {x - 1} \right)^2 }}dx} = 2\int {\left( {x - 1} \right)^{ - 2} d\left( {x - 1} \right)} [/tex]
     
  6. Nov 28, 2005 #5
    I'm a little confused, would it be..

    [tex]2[\frac{(x-1)^-1}{-1}][/tex] with limits 2 and 1? Actually that wouldn't work because of 0^-1, is there any way to get rid of the -1?
     
    Last edited: Nov 28, 2005
  7. Nov 28, 2005 #6

    TD

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    You integrated that second part correctly, but the limits just stay the same as in your original problem, being 2 and 3.
     
  8. Nov 28, 2005 #7
    Therefore the answer would be (3+-1)-(2+-2) = 2?
     
  9. Nov 28, 2005 #8

    TD

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    2 is correct, if you're referring to the entire (initial) integral.
     
  10. Nov 28, 2005 #9
    Yes, thank you for your help.
     
  11. Nov 28, 2005 #10

    TD

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    No problem :smile:
     
  12. Nov 28, 2005 #11
    Oh by the way, is it correct in saying (X+1)^2 is the same as (X+1)(X-1)
     
  13. Nov 28, 2005 #12

    TD

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    Certainly not, you can see that by working out both expressions.

    Of course, it's easy to see that (x+1)² = (x+1)(x+1) and that's not the same as (x+1)(x-1).
    Perhaps you're confused with the identity a²-b² = (a-b)(a+b) with a = x and b = 1.
     
  14. Nov 28, 2005 #13
    Yes I was getting mixed up with the difference of two squares because I am trying to solve [tex]\int_{0}^{1}_\frac{x^2+2x}{(x+1)^2}dx[/tex]
    This led me to[tex]\int_{0}^{1}_\frac{(x-1)^2-1}{(x+1)^2}dx[/tex] but I have completed the square wrongly and it should be [tex](x+1)^2-1[/tex] which was leading be astray a bit :)
     
    Last edited: Nov 28, 2005
  15. Nov 28, 2005 #14

    TD

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    Ok, but were you able to do it or are you stuck somewhere?
     
  16. Nov 28, 2005 #15
    Yep thanks to your help earlier I have completed the question. I got the answer as 0.5
     
  17. Nov 28, 2005 #16

    TD

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    That is correct :smile:
     
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