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Thread: Integrals involving exponential

  1. #1
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    Exclamation Integrals involving exponential

    Question:
    $\displaystyle \int \frac{dx}{1+e^x} $

    How to solve this ..?

    Please answer in steps thank you
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by mj.alawami View Post
    Question:
    $\displaystyle \int \frac{dx}{1+e^x} $

    How to solve this ..?

    Please answer in steps thank you
    $\displaystyle u = 1+e^x$

    $\displaystyle du = e^x dx$

    $\displaystyle dx = (u-1)\, du$

    $\displaystyle \int \frac{(u-1)}{u} \, du = \int 1\,du - \int \frac{1}{u}\, du$
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  3. #3
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    Try this...

    Let $\displaystyle e^x$=t

    x=ln(t)

    dx=(1/t)*dt

    I=$\displaystyle \int 1/(1+e^x) dx = \int [(1/(t+1)*(1/t)] dt$

    I=$\displaystyle \int$ {(1/t)-[1/(t+1)]} dt

    $\displaystyle I=ln(t)-ln(t+1)=ln( e^x)-ln( e^x+1)$

    I=ln[ex/( ex+1)]+C

    I found this proof on the web but verified it in calculation. Seems right to me, hope this helps.
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    $\displaystyle u = 1+e^x$

    $\displaystyle du = e^x dx$

    $\displaystyle dx = (u-1)\, du$

    $\displaystyle \int \frac{(u-1)}{u} \, du = \int 1\,du - \int \frac{1}{u}\, du$

    I still didnt understad , Sorry
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by mj.alawami View Post
    I still didnt understad , Sorry
    I used integration by substituion.

    Let $\displaystyle u = 1+e^x $ - this was chosen to make the denominator easier

    When we differentiate u wrt x we get $\displaystyle du = e^x \, dx$ which is obtained through normal rules.

    To be able to substitute our du term into the equation we need to isolate dx: $\displaystyle dx = \frac{du}{e^x}$. However that isn't very helpful so to get that e^x out the way rearrange the original substitution

    $\displaystyle e^x = u-1$

    $\displaystyle dx = \frac{du}{u-1}$

    (that was a mistake I made in my previous post. It should read the above)

    Putting this back into the integral for dx:

    $\displaystyle \int \frac{du}{u(u-1)} = \int \frac{du}{u^2-u}$
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