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Math Help - New to Integration (and new to forum)

  1. #1
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    New to Integration (and new to forum)

    Hi guys, calculus is not my strong point and I'm having a bit of difficulty trying to integrate these problems:

    integrate: x(x-2)/(x-1)^3
    int: x/(9-x)^1/2

    Can I use u substitution here?


    and also I'm not sure how to go about these problems, where I have to find the area under the curve bounded the given numbers:

    1/xlnx where b=e^2 and a=e

    x-1/(x+1) where b=1 and a=0

    I would really appreciate it!

    <3
    from a humanities major
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  2. #2
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    The key is to exterminate the root, that is what annoys us, so for your first problem, put t=\sqrt[3]{x-1} written backwards as t^3=x-1, and then compute \frac{dt}{dx}.

    In the same fashion, can you compute your second integral?
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  3. #3
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    I'm not sure what you mean since my first problem isn't a root, it's cubed. am I supposed to take a cubed root there?
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  4. #4
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    oh wait, do you mean x(x-2)^(1/3) / (x-1)
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  5. #5
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    Ohh!! yes, you're right! ahhh well, it's even simple! put t=x-1 and use that \frac{a+b}c=\frac ac+\frac bc.

    anyway, what I did before still works for your second problem.
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  6. #6
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    thanks for replying so quickly!

    I suppose I don't understand what you mean by the a+b/c = a/c +b/c. can I still do that when x is multiplied by (x-2)? Maybe I'm missing a step here.


    x(x-2) x x-2
    ----------- ?=? ---------- + ---------
    (x-1)^3 (x-1)^3 (x-1)^3




    i'm sorry if i'm a little bit slow. I'm taking calc 2, and haven't taken calculus since high school.
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  7. #7
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    Quote Originally Posted by PandaNomium View Post

    integrate: x(x-2)/(x-1)^3
    Is this your integral right? so then put t=x-1, what do you get?
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  8. #8
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    (t+1)(t-1)/ t^3 ?
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  9. #9
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    yes, now that is \frac{t^2-1}{t^3}=\frac1t-\frac1{t^3}, and those are easy to integrate.
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  10. #10
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    oooh.

    is it ln(x-1) + 1/2(x-1)^2 + C?
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  11. #11
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    correct! (i see you added the constant! well done!)
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  12. #12
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    thanks my brain is a little fuzzy... :P
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  13. #13
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    so for the second problem I set t^2=9-x^2. Then when i derive it, it becomes t=x?

    I'm kind of stuck there.
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  14. #14
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    no, it's just t^2=9-x.
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  15. #15
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    oh i'm sorry, i wrote the problem wrong:

    it's suppose to be x/sqrt(9-x^2)
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