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Math Help - logarithm question need a slight help

  1. #1
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    logarithm question need a slight help

    logarithm question need a slight help-1111111.jpgthis i tried to integrate by u substitution U=e^x+1 and then U-1/U after this i get U-lnU i dont know what do can anyone hint or help me solving it
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  2. #2
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    Re: logarithm question need a slight help

    Your substitution is good:

    u=e^x+1\,\therefore\,du=e^x\,dx giving us:

    \int_2^{e+1} \frac{1}{u}\,du

    Can you proceed from here?

    Incidentally, the given answer is not quite correct.

    By the way, I would post integration problems in the calculus forum.
    Last edited by MarkFL; December 15th 2012 at 03:09 AM.
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    Re: logarithm question need a slight help

    not really u changed the question while i took it from a very reliable source
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    Re: logarithm question need a slight help

    Quote Originally Posted by abdulrehmanshah View Post
    not really u changed the question while i took it from a very reliable source
    This is what you posted: \int_0^1 {\frac{{e^x }}{{1 + e^x }}dx}  = \mathop {\left. {\ln \left( {1 + e^x } \right)} \right|}\nolimits_{x = 0}^{x = 1}

    And the give answer is wrong.
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  5. #5
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    Re: logarithm question need a slight help

    Quote Originally Posted by abdulrehmanshah View Post
    not really u changed the question while i took it from a very reliable source
    \int_0^1 \frac{e^x}{e^x+1} \, dx = \left[\ln(e^x+1)\right]_0^1 = \ln(e+1) - \ln(2) = \ln\left(\frac{e+1}{2}\right)

    I agree with MarkFL2. I'd check that "source" again if I were you ...
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    Re: logarithm question need a slight help

    ok the answer can be wrong but how can u change the limits provided by question can anyone help solving
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    Re: logarithm question need a slight help

    Quote Originally Posted by abdulrehmanshah View Post
    ok the answer can be wrong but how can u change the limits provided by question can anyone help solving
    when a substitution is done, one may "reset" the limits of integration from x-values to u-values ...

    u = 1 + e^x

    for x = 0 , u = 1 + e^0 = 2

    for x = 1 , u = e^1 + 1 = e + 1

    therefore ...

    \int_0^1 \frac{e^x}{1+e^x} \, dx = \int_2^{e+1} \frac{1}{u} \, du = \left[\ln{u}\right]_2^{e+1} = \ln(e+1) - \ln(2) = \ln\left(\frac{e+1}{2}\right)
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    Re: logarithm question need a slight help

    this is the correct answer and as stated by my book so what is that is wrong about this question
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    Re: logarithm question need a slight help

    Quote Originally Posted by abdulrehmanshah View Post
    this is the correct answer and as stated by my book so what is that is wrong about this question
    LOOK at your "answer" again ...

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  10. #10
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    Re: logarithm question need a slight help

    where did the e^x in the numerator go i know that lnu= 1/u but where did e^x go?
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  11. #11
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    Re: logarithm question need a slight help

    ya i wrote it wrongly my bad the answer u got is the answer i have in the back of my note book can u please explain e^x part
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    Re: logarithm question need a slight help

    Quote Originally Posted by abdulrehmanshah View Post
    this is the correct answer and as stated by my book so what is that is wrong about this question
    I don't know what your book gave but you posted the answer as ln\left(\frac{1+ e^x}{2}\right). Since this is a definite integral there should be no "x" in the answer. The correct answer is, as skeeter said, ln\left(\frac{1+ e}{2}\right)
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  13. #13
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    Re: logarithm question need a slight help

    \int_0^1 \frac{{\color{red}e^x}}{1+e^x} \, {\color{red}dx} = \int_2^{1+e} \frac{{\color{red}du}}{u}


    u = 1 + e^x

    {\color{red}du = e^x \, dx}
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