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U-substitution problem

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    ##\displaystyle \int \frac{\ln x}{x(1 + \ln x)} dx##

    2. Relevant equations

    3. The attempt at a solution

    Let ##u = 1 + \ln x##, then ##\ln x = u - 1##
    ##\displaystyle du = \frac{1}{x}dx##
    Thus, ##\displaystyle \int \frac{\ln x}{x(1 + \ln x)} dx = \int \frac{u - 1}{u} du= \int 1 - u^{-1} du = u - \ln u + C = 1 + \ln x - \ln \left | \ln x + 1\right | + C##

    However, this is the wrong answer. What am I doing wrong?
  2. jcsd
  3. Nov 30, 2015 #2

    Ray Vickson

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    Why do you think it is wrong?
  4. Nov 30, 2015 #3
    Because my answer book says that the answer should be ##\ln x - \ln | \ln x + 1 | + C## I have a 1 in there, which is where the answer differs.
  5. Nov 30, 2015 #4


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    Is not C+1 a constant?
  6. Nov 30, 2015 #5
    Does that mean as my final answer I should redefine C + 1 as just C, since they are both just constants? Why did I get a 1 in the first place, while other ways of obtaining the solution don't have the 1? Such as letting u be just lnx instead of lnx + 1?
  7. Dec 1, 2015 #6


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    When you integrate, you find one result, one primitive of the function you integrate.
    Now, if the function ##f## is the the primitive you found, ##f+C## (##C## any constant) will also be a primitive. So yes, you can remove the +1, or leave it, both results are correct. Removing the +1 makes the answer somewhat more elegant, but there is nothing more to it than that.
  8. Dec 1, 2015 #7

    Ray Vickson

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    You got the '1' in the first place because YOU chose to define u as 1 + ln(x).

    Anyway, why are you obsessing on this issue? It as absolutely NO importance: the constant of integration is totally arbitrary, and can be called 147.262 + K instead of C, if that is what you prefer. It can be called anything you want, and if somebody else chooses to call it something different from what you choose to call it, that is OK, too. If you use a computer algebra system to do the indefinite integral, it might not a constant of integration (but it might include your "1" if it happened to use your change-of-variable method).
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