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Math Help - Clarification on Definite Integrals

  1. #1
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    Clarification on Definite Integrals

    Hello, and thanks in advance for your help.

    The problem I'm facing is to integrate by parts the following function:

    lnx/x^2 with the interval of integration being from 1 to 2.

    I don't think I need a full solution, it's just that after I do the integration by parts method, I'm left with no integrals, and this makes me concerned. Is the answer to this problem simply my simplified 1/3x^3lnx - 1/9x^3 or is it this simplification with the values 2 inputted for x subtracted from it with 1 inputted for x (like shown below)?

    [1/3*2^3*ln2 - 1/9*2^3] - [1/3*1^3*ln1 - 1/9*1^3]

    Thanks again for the assistance...I think it's just a matter of not understanding how the interval of integration given at the start of the problem relates to my solution.
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  2. #2
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    integrate ln(x)/x^2 - Wolfram|Alpha

    click show steps on the right to see the integration by parts
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by NBrunk View Post
    Hello, and thanks in advance for your help.

    The problem I'm facing is to integrate by parts the following function:

    lnx/x^2 with the interval of integration being from 1 to 2.

    I don't think I need a full solution, it's just that after I do the integration by parts method, I'm left with no integrals, and this makes me concerned. Is the answer to this problem simply my simplified 1/3x^3lnx - 1/9x^3 or is it this simplification with the values 2 inputted for x subtracted from it with 1 inputted for x (like shown below)?

    [1/3*2^3*ln2 - 1/9*2^3] - [1/3*1^3*ln1 - 1/9*1^3]

    Thanks again for the assistance...I think it's just a matter of not understanding how the interval of integration given at the start of the problem relates to my solution.
    Try letting \ln(x)=z.
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  4. #4
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    Thanks for the link, it showed how wrong my solution was...took dv = simply x^2 rather than x^-2 and it threw me off.

    However, the link shows the solution to the indefinite integral, and I'm still not sure how to reflect the definite interval of integration of 1 to 2 in my answer...although I'm much closer!
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Okay listen (actually read); we want to compute \int_1^2\frac{\ln x}{x^2}\,dx.

    You already spotted that this needs to be integrated by parts, now the trick here, it's to express \ln x or \frac1{x^2} as a function being differentiated, but you'll probably say "I don't get your point." It's simple, let's think: do we have a function whose derivative gives me \ln x ? Yes, it does exist, but, do you know it that fast so that you can tell me what it is? - No, so let's think on \frac1{x^2}, we can immediately see that \left(-\frac1x\right)'=\frac1{x^2}, so we have \int_{1}^{2}{\frac{\ln x}{x^{2}}\,dx}=\int_{1}^{2}{\left( -\frac{1}{x} \right)'\ln (x)\,dx}. This will be equal to the following: multiply -\frac{1}{x} and \ln x evaluated for 1\le x\le2 and the remaining integral will be -\int_1^2\frac1{x}(\ln x)'\,dx, so our original integral becomes \left. -\frac{1}{x}\ln x \right|_{1}^{2}+\int_{1}^{2}{\frac{1}{x}\left( \ln x \right)'\,dx}=-\frac{1}{2}\ln 2+\int_{1}^{2}{\frac{dx}{x^{2}}}=-\frac{1}{2}\ln 2+\left( \left. -\frac{1}{x} \right|_{1}^{2} \right).

    So our integral equals -\frac{1}{2}\ln 2+1-\frac{1}{2}=\frac{1-\ln 2}{2}.
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  6. #6
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    Correction and Restatement of Question

    Pending Correction
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  7. #7
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    1) let ln x=t
    2) int(e^t.t)dt
    3) by parts integration
    4) use ILATE rule
    5) ans:e^t(t-1)
    6) ans in terms of x:x(ln x-1)
    7) put the limits
    please comment..........
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