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Clarification needed: u-sub requires combining 2 constants

  1. Jun 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫-x/(x+1) dx


    2. Relevant equations



    3. The attempt at a solution

    I do a U-sub

    u=x+1
    du=dx
    x=u-1

    -∫(u-1)/u du

    -∫[u/u - 1/u] du

    -∫1 - 1/u du

    -[∫1du - ∫1\u du]

    -[u - ln(u) + C]

    ln(u)-u+C

    ln(x+1)-(x+1)+C

    ln(x+1)-x-1+C

    I combine -1 and C into a single C to get the final answer

    ln(x+1) - x + C


    Wolfram says the answer is correct, but its method is slightly different. It uses long division at the outset to go from -∫x/(x+1) dx → -∫1 - 1/(x+1) dx before seperating into 2 integrals, and doing a u-sub on the second.


    My answer ends up the same, but getting that extra -1 from substituting the (x+1) in for u is making me nervous for some reason, even though I ultimately combine it with the constant. Is my process equally correct (I think it is) or did I just get lucky with the solution?

    Thanks
     
    Last edited: Jun 24, 2013
  2. jcsd
  3. Jun 24, 2013 #2

    Mark44

    Staff: Mentor

    Either approach can be taken to produce the antiderivative. Many times it is the case that more than one technique can be used.

    Two comments:
    1. You should connect the expressions you have with =.
    2. In these lines --
    ln(x+1)-x-1+C
    ln(x+1) - x + C
    -- C on the 2nd line is different from C on the first. What people usually do is write a different constant (e.g., C') to indicate that the new constant actually is different from the previous one.
     
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