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Thread: Integrating e^x

  1. #1
    Nas
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    Exclamation Integrating e^x

    I am unsure on how to calculate the following 2 integrals

    a) [1/(x^2)]*[e^(2/x)], with respect to xand
    b) 1/(e^x + 1), with respect to x

    Help with this will be greatly appreciated.
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  2. #2
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    Quote Originally Posted by Nas View Post
    I am unsure on how to calculate the following 2 integrals

    a) [1/(x^2)]*[e^(2/x)], with respect to xand
    b) 1/(e^x + 1), with respect to x

    Help with this will be greatly appreciated.
    Here you have

    a) $\displaystyle \int{\left(\frac{1}{x^2}\right)e^{\frac{2}{x}}\,dx } = \int{e^{2x^{-1}}\,x^{-2}\,dx}$

    $\displaystyle = -\frac{1}{2}\int{e^{2x^{-1}}(-2x^{-2})\,dx}$.

    Now make the substitution $\displaystyle u = 2x^{-1}$ so that $\displaystyle du = -2x^{-2}\,dx$ and the integral becomes

    $\displaystyle -\frac{1}{2}\int{e^u\,du}$

    $\displaystyle = -\frac{1}{2}e^u + C$

    $\displaystyle = -\frac{1}{2}e^{2x^{-1}} + C$

    $\displaystyle =-\frac{1}{2}e^{\frac{2}{x}} + C$.
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  3. #3
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    Quote Originally Posted by Nas View Post
    I am unsure on how to calculate the following 2 integrals

    a) [1/(x^2)]*[e^(2/x)], with respect to xand
    b) 1/(e^x + 1), with respect to x

    Help with this will be greatly appreciated.
    b) $\displaystyle \int{\frac{1}{e^x + 1}\,dx} = \int{\frac{e^x}{e^x(e^x + 1)}\,dx}$

    $\displaystyle = \int{\left[\frac{1}{e^x(e^x + 1)}\right]e^x\,dx}$.

    Now make the substitution $\displaystyle u = e^x + 1$ so that $\displaystyle e^x = u - 1$ and $\displaystyle du = e^x\,dx$.

    The integral becomes

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

    Now solve this using partial fractions.
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  4. #4
    Nas
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    would the answer for part b be x-log(e^x+1)?
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    1. Let $\displaystyle u = \dfrac{2}{x}$, then $\displaystyle \dfrac{du}{dx} = \dfrac{-2}{x^2} \Rightarrow {dx} = -\dfrac{x^2{du}}{2}$. Therefore $\displaystyle \int\dfrac{1}{x^2}e^{\frac{2}{x}}\;{dx} = -\int\left(\dfrac{e^u}{x^2}\right)\left(\dfrac{x^2} {2}\right)\;{du} = ...$
    2. Let $\displaystyle u = e^x+1$, then $\displaystyle \dfrac{du}{dx} = e^x \Rightarrow {dx} = \dfrac{du}{e^x}.[/Math] Therefore [Math]\int\dfrac{1}{e^x+1}\;{dx} = \int\dfrac{1}{u(u-1)}\;{du} = ... $
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  6. #6
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    Another solution for 2:

    $\displaystyle \int \frac{dx}{e^x+1} = \int \frac{1}{e^x+1} \, \frac{e^{-x}}{e^{-x}} \, dx$

    $\displaystyle =\int \frac{e^{-x}}{e^{-x}+1} \, dx$

    Let $\displaystyle u=e^{-x}+1$, to get :

    $\displaystyle - \int \frac{du}{u} = -ln|u|+C = -ln(e^{-x}+1) + C$

    $\displaystyle =ln \left( \frac{1}{e^{-x}+1} \right) + C$
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  7. #7
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    Hello, Nas!

    The General beat me to the solution . . . *sigh*


    $\displaystyle (b)\;\;\int\frac{dx}{e^x+1} $

    I have an equivalent answer: .$\displaystyle -\ln\left(1 + e^{-x}\right) + C$


    And this answer can be simplified beyond all recognition . . .


    $\displaystyle -\ln\left(1 + \frac{1}{e^x}\right) + C\;=\;-\ln\left(\frac{e^x+1}{e^x}\right) + C$

    . . . . . . . . . . . . $\displaystyle =\;-\ln(e^x+1) + \ln(e^x) + C$

    . . . . . . . . . . . . $\displaystyle =\;-\ln(e^x+1) + x\underbrace{\ln(e)}_{\text{This is 1}} + C$

    . . . . . . . . . . . . $\displaystyle =\;x - \ln(e^x+1) + C$

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  8. #8
    Nas
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    Thanks for all the help.
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