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Math Help - what did i do wrong? Integral problem:

  1. #1
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    what did i do wrong? Integral problem:

    Am i doing this problem wrong?
    Evaluate the integral
    <br />
\int (e^(2x)) / (e^x  - 2)) dx <-- that is e^(2x)
    i expanded the problem to:
    \int(e^x * e^x *  (1/ (e^x  - 2)) dx

    i picked u to be e^x - 2
    u = e^x  - 2, e^x = u + 2
    du = e^x
    rewrite to:
    \int((u + 2) *  1/ (u) dx

    multiply all out to
    \int 1 + 2*(1/u) dx

    integrate and get:
    u + 2ln|u| + C
    (e^x - 2) + 2ln|e^x - 2|+C

    but the answer says its:
    e^x + 2ln|e^x - 2|+C

    what did i do wrong?
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  2. #2
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    you didn't do anything wrong. your answer is correct. notice that the -2 from (e^x - 2) can be combined with the constant C to produce another arbitrary constant. once you do that, your answer will look just like the other answer.
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  3. #3
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    you can combine it with the constant? Hmm never done that before, thanks.
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  4. #4
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    Hello, micro31337!

    You did nothing wrong . . .


    Integrate and get: . (e^x - 2) + 2\ln|e^x - 2|+C

    but the answer says it's: . e^x + 2\ln|e^x - 2|+C

    \text{You have: }\;e^x - 2 \;+\; 2\ln|e^x-2| \;+\; C \;\;=\;\;e^x \;+\; 2\ln|e^x-2| \;+\; \underbrace{C - 2}
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
    This is an arbitrary constant

    Therefore: . e^x + 2\ln|e^x-2| + C

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