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Indefinite Integral of (1/x^2)

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Hello, first of all I would like to apologize for the fact that this question is extremely trivial compared to the other questions being asked. I have a improper integral problem, and the entire problem itself is not relevant, because I understand everything in it except for one thing. One step in the problem requires finding the indefinite integral of (1/x^2). The example in the book tells me that the answer is (-1/x), and it says that the answer is obtained by using the reverse power rule, but I just can’t see how they got that answer. I have a feeling that it is something very simple, and that I’ve forgotten some subtle detail.



    2. Relevant equations






    3. The attempt at a solution


    Here’s how I tried to solve it:

    (1/x^2) has an overall power of one, right? I increased the power of the whole thing by one, which is (1/x^2)^2, and I divided the whole thing by two, which is the same thing as multiplying by ½. So…

    (½)(1/x^2)^2

    You can already see that this is not going to give an answer of (-1/x). Could someone please help me with this?
     
  2. jcsd
  3. Jun 8, 2009 #2

    Avodyne

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    Science Advisor

    [tex]\int x^n\,dx = {1\over n+1}x^{n+1} + C[/tex]
    In your case, [itex]n=-2[/itex].
     
  4. Jun 8, 2009 #3
    Hmmm....okay, I'll buy that. But why is n negative 2? If I'm raising the whole thing to the power of 2, why isn't n just positive 2? Thank you!
     
  5. Jun 8, 2009 #4
    [tex]
    \frac{1}{x^n} = x^{-n}
    [/tex]

    A basic property of exponents.
     
  6. Jun 8, 2009 #5

    Cyosis

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    Homework Helper

    Raising the entire expression to the power of 2 would give you [itex](1/x^2)^2=1/x^4=x^{-4}[/itex].
     
  7. Jun 8, 2009 #6

    Mark44

    Staff: Mentor

    To expand on what kbaumen wrote, and relative to your problem,
    [tex]\frac{1}{x^2} = x^{-2}[/tex]
     
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