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Math Help - integration by substitution

  1. #1
    Member Jskid's Avatar
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    integration by substitution

    Question: evaluate the indefinite integral by using substitution
    \int \frac{e^x+1}{e^x}
    The answer key has x-e^{-x}+C

    I tried the substitutions

    1. u=e^x\Rightarrow du=e^x dx
    2. u=e^x+1\Rightarrow du=e^x dx

    1. doesn't really work so I think 2. is a better bet. I think I need to somehow get du=\frac{dx}{e^x} but I don't know how.
    Last edited by Jskid; December 23rd 2010 at 12:39 PM. Reason: fixed Latex on answer, -x is an exponent
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  2. #2
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    \displaystyle \frac{e^{x}+1}{e^x} = \frac{e^x}{e^x}+\frac{1}{e^x} = 1+\frac{1}{e^x} = 1+e^{-x}.

    Quote Originally Posted by Jskid View Post
    1. doesn't really work so I think 2. is a better bet. I think I need to somehow get du=\frac{dx}{e^x} but I don't know how.
    Both subs should work. The first one gives:

    \displaystyle \frac{du}{dx} = e^x \Rightarrow dx = \frac{du}{e^x} so \displaystyle \int\frac{e^x+1}{e^x}\;{dx} = \int\frac{e^x+1}{e^x\cdot e^x}}\;{du} = \int\frac{u+1}{u^2}\;{du} = ...
    Last edited by TheCoffeeMachine; December 23rd 2010 at 01:05 PM.
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  3. #3
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    Hello, Jskid!

    \text{Evaluate the inde{f}inite integral by using substitution:}

    . . \displaystyle \int \frac{e^x+1}{e^x}\,dx

    Here's your second substitution . . .

    Let u \:=\:e^x+1 \quad\Rightarrow\quad e^x \:=\:u-1
    . . . . x \:=\:\ln(u-1) \quad\Rightarrow\quad dx \:=\:\dfrac{du}{u-1}

    Substitute: . \displaystyle \int\frac{u}{u-1}\cdot\frac{du}{u-1} \;=\;\int\frac{u\,du}{(u-1)^2}

    Partial fractions: . \displaystyle \int\left(\frac{1}{u-1} + \frac{1}{(u-1)^2}\right)du \;=\;\int\frac{du}{u-1} + \int(u-1)^{-2}du

    Integrate: . \ln|u-1| - (u-1)^{-1} + C

    Back-substitute: . \ln|(e^x+1)-1| - \dfrac{1}{(e^x+1)-1} + C

    . . . . . . . . . . \displaystyle =\;\ln|e^x| - \frac{1}{e^x} + C

    . . . . . . . . . . =\;x - e^{-x} + C

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  4. #4
    Member Jskid's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    \displaystyle \frac{e^{x}+1}{e^x} = \frac{e^x}{e^x}+\frac{1}{e^x} = 1+\frac{1}{e^x} = 1+e^{-x}.
    I'm with you so far but I don't know how to integrate e^{-x}
    let u=e^{-x} \Rightarrow du=-e^{-x} dx what next?

    @Soroban the textbook hasn't introducted partial fractions yet. Is there a way to solve this that doesn't require them?
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  5. #5
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    e^(i*pi)'s Avatar
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    Think about differentiation in reverse... You know the derivative of e^-x so work backwards

    For reference  \displaystyle \int e^{ax} = \dfrac{1}{a}e^{ax}
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  6. #6
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    Quote Originally Posted by Jskid View Post
    I'm with you so far but I don't know how to integrate e^{-x}
    let u=e^{-x} \Rightarrow du=-e^{-x} dx what next?
    @Soroban the textbook hasn't introducted partial fractions yet. Is there a way to solve this that doesn't require them?
    This problem has nothing to do with partial fractions.
    Nothing more than basic, basic algebra tell us that \dfrac{e^x+1}{e^x}=1+e^{-x}.

    Can you differentiate -e^{-x}~?
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    For reference  \displaystyle \int e^{ax} = \dfrac{1}{a}e^{ax}
    This is not apperant to me. It looks simple but I don't actually know how it works, could someone elaborate?
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  8. #8
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    Quote Originally Posted by Jskid View Post
    This is not apperant to me. It looks simple but I don't actually know how it works, could someone elaborate?
    This a general result, commit it to memory, if you want a better understanding apply the chain rule to find the derivative of \dfrac{1}{a}e^{ax}
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  9. #9
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    Use the substitution ax = u \Rightarrow dx = \frac{du}{a}.
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  10. #10
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    Ok, I'll try to explain...
    Quote Originally Posted by Jskid View Post
    I'm with you so far but I don't know how to integrate e^{-x}
    let u=e^{-x} \Rightarrow du=-e^{-x} dx what next?
    Don't solve for du, solve for dx: we have {dx} = -\frac{1}{e^{-x}}\;{du}.

    Your integral was \int e^{-x} \;{dx}, replacing the dx with -\frac{1}{e^{-x}}\;{du}, we have:

    \int e^{-x} \;{dx} = \int e^{-x}\left(-\frac{1}{e^{-x}}\;{du}\right) = -\int\frac{e^{-x}}{e^{-x}}\;{du} = -\int~ 1~ \;{du} = -u+k

    But u = e^{-x}, so \int e^{-x} \;{dx} = -e^{-x}+k.

    Do you understand now?
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  11. #11
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    Also, just in case a picture helps...



    ... where (as usual) ...



    represents the chain rule for differentiation - straight continuous lines differentiating down (integrating up) with respect to x, the straight dashed line similarly but with respect to the dashed balloon expression.

    PS: I should have emphasised (and after Plato's comment below now can't resist)... read the diagram downwards to differentiate...



    and upwards to integrate...


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; December 23rd 2010 at 03:53 PM. Reason: PS
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  12. #12
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    Quote Originally Posted by Defunkt View Post
    Use the substitution ax = u \Rightarrow dx = \frac{du}{a}.
    I think that the method of u-substitution is most deleterious, non-productive idea that matheducators have foisted off on poor students honestly trying to learn to do anti-differentiation. It gives a false sense of understanding of the actual process. Now I do agree with the philosopher of mathematics Michael Resnik (I admit being a friend), he is known as a structurelist: mathematics is the science of pattern.

    To find an anti-derivative, one needs to look at the pattern of derivatives.
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  13. #13
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    So you would rather not teach substitution at all? [off topic, so I won't continue]
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  14. #14
    Member Jskid's Avatar
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    Thanks people I get it now.
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