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Homework Help: Evaluating a Definite Integral

  1. Jul 12, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫ln(x+1)/(x+1)2dx from a=1, b=2

    (Answer is -ln3/3+ln2/2+1/6)

    2. Relevant equations

    3. The attempt at a solution
    Let u = (x+1), du=dx

    Let f=lnu , df = 1/udu, g=-1/u, dg=1/u2du

    =[-lnu+1/u] 1->2

    Using Fundamental Theorem of Calculus plug in a,b


    What did I miss in order for me not to get 1/6 in the answer? Help would be great!

  2. jcsd
  3. Jul 12, 2012 #2


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    When you used a change of variable right at the beginning of u= x+1, you have to evaluate your limits of integration for the new variable. Ie after using parts, you should integrate between 3 and 2.
  4. Jul 12, 2012 #3


    Staff: Mentor

    No, you don't have to evaluate at the new limits when you do a change of variable. If you "undo" the substitution, you can use the original limits.

    Let f=lnu , df = 1/udu, g=-1/u, dg=1/u2du

    You are missing a u in the denominator of the first term, so it should be -ln(u)/u.

    Undo the substitution to get -ln(x + 1)/(x+1) + 1/(x + 1) and evaluate at 2 and 1, as usual, and you'll get the answer you showed.
  5. Jul 12, 2012 #4
    Thanks for the replies!

    I tried out both methods but did not have any luck. With the first method provided by CAF123. I was unable to get an answer close to what it should be. With Mark's method. I received an answer of -ln3/3+ln2/2+2. I still cannot find where I am going wrong to not get 1/6.
  6. Jul 13, 2012 #5


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    What is [itex]\displaystyle \int \frac{1}{u^2}\,du\ ?[/itex]

    (You have a sign error.)

    Last edited: Jul 13, 2012
  7. Jul 13, 2012 #6


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    In the end, you should be integrating -ln(u)/u -1/u with upper limit 3, lower limit 2.
    Or as Mark44 said, integrate -ln(x+1)/(x+1) -1/(x+1) with upper limit 2, lower limit 1.
    (To Mark44: I understood this perfectly, I think I just misinterpreted what the last line was in the OP's attempt).
    Many thanks.
  8. Jul 13, 2012 #7
    How come? isn't that suppose to be ∫gdf? In which, pulling out the negative will make the integral become positive and 1/u^2du.
  9. Jul 13, 2012 #8


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    Correct, but the integral of 1/u^2 is -1/u
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