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Math Help - Substitution and Integration by Parts Problem, I keep getting it almost right!

  1. #1
    s3a
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    Substitution and Integration by Parts Problem, I keep getting it almost right!

    The question (and Wolfram's Answer) is: integral ln(2x+1 - Wolfram|Alpha) (Sorry, I still don't know how to use laTex)

    however, I, Wolfram and my textbook each have different answers. So, my first though was that it was just different forms but then I plugged in x = 5 and get different answers with each equation. Maybe the difference is due to the constant not being used?

    Here is the answer on the back of the book:
    1/2 * (2x+1) * ln(2x+1) - x + C

    My work is attached.

    Any help would be greatly appreciated as always!
    Thanks in advance!
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  2. #2
    o_O
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    If you multiply Wolfram's answer out:

    \frac{1}{2} (2 x+1) (\ln(2 x+1)-1)+K

    = \frac{1}{2}(2x+1)\ln (2x+1) - \frac{1}{2}(2x+1) + K

    =\frac{1}{2}(2x+1)\ln (2x+1) - x \underbrace{-\  \frac{1}{2} + K}_{\text{Call this C}}

     =\frac{1}{2}(2x+1)\ln (2x+1) - x + C

    Happier now ? In other words, yes the two answers simply differ by a constant.
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  3. #3
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    Hello s3a
    Quote Originally Posted by s3a View Post
    The question (and Wolfram's Answer) is: integral ln(2x+1 - Wolfram|Alpha) (Sorry, I still don't know how to use laTex)

    however, I, Wolfram and my textbook each have different answers. So, my first though was that it was just different forms but then I plugged in x = 5 and get different answers with each equation. Maybe the difference is due to the constant not being used?

    Here is the answer on the back of the book:
    1/2 * (2x+1) * ln(2x+1) - x + C

    My work is attached.

    Any help would be greatly appreciated as always!
    Thanks in advance!
    I think the book and Wolfram are both right. You have lost a factor of \tfrac12 on the last line of your working, before the final \ln(2x+1) term.

    Any other apparent differences in the answers are down to (a) re-arranging the expression (b) the constant of integration.

    Grandad
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  4. #4
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    Note that

    \frac{d}{dx} f(x) \ln (f(x)) = (1 + \ln(f(x)) \cdot f'(x)).

    Hence,

    \frac{d}{dx} (2x + 1) \ln (2x + 1) = (1 + \ln(2x + 1)) \cdot 2

    \int (1 + \ln(2x + 1)) \cdot 2 dx = 2 \int (1 + \ln(2x + 1))dx = 2 \int dx + 2 \int \ln(2x + 1) dx

    (2x + 1) \ln (2x + 1) = 2 \int dx + 2 \int \ln(2x + 1) dx

    (2x + 1) \ln (2x + 1) - 2 \int dx = 2 \int \ln(2x + 1) dx

    \frac{1}{2}\left((2x + 1) \ln (2x + 1) - 2 \int dx \right) = \int \ln(2x + 1) dx
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