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Math Help - Evaluating integrals on exponential functions with natural logs.

  1. #1
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    Evaluating integrals on exponential functions with natural logs.

    ∫(e-x/(1+e-x))
    So I used substituion, u = 1 + e-x, and -du = e-x dx
    There it turns it into -∫(du/u), which equals -ln(1+e-x) + C
    Then my partner told me the next step was ln(ex/(ex+1))
    After that, he got it into x - ln (ex + 1) + C, which is the correct answer.
    And I was confused on that part since it turned into something like a fraction. The part where it goes e x/ex+1 Can someone explain?

    Another is ∫ex√(1-ex)dx
    I did u = 1-ex, -du = ex dx
    I plugged it in...
    ∫-du√(u) = -∫du(u1/2)
    Not really sure if I'm doing this next step correctly... but I did that it = u1/2
    Plugged in u: (1-ex)1/2

    But the answer was: -2/3(1-ex)3/2 + C

    I'm a bit confused now... since I'm integrating the 'e' along...
    Can someone give a piece of advice or explain please?

    Thanks
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  2. #2
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    Re: Evaluating integrals on exponential functions with natural logs.

    1. -\ln\left(1+\frac{1}{e^x}\right)=-\ln\left(\frac{e^x+1}{e^x}\right)=-\ln\left(\frac{e^x}{e^x+1}\right)^{-1}=\ln \frac{e^x}{e^x+1}

    2. \int \sqrt{u}\, du =\frac{u^{3/2}}{3/2}+C
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  3. #3
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    Re: Evaluating integrals on exponential functions with natural logs.

    Quote Originally Posted by princeps View Post
    1. -\ln\left(1+\frac{1}{e^x}\right)=-\ln\left(\frac{e^x+1}{e^x}\right)=-\ln\left(\frac{e^x}{e^x+1}\right)^{-1}=\ln \frac{e^x}{e^x+1}

    2. \int \sqrt{u}\, du =\frac{u^{3/2}}{3/2}+C
    Oh wow.
    Now that makes more sense!

    I forgot to see that it had a negative exponent, which makes it goes down into the denominator, so the only way to convert the 1 to have them both at the same denominator is by having it ex as well!

    While for the second one. I think I remember now! So basically after you integrate it, the exponent goes up by 1 and you would divide it by the exponent.

    Thanks for helping me and reminding me of this stuff!
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