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Math Help - help please

  1. #1
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    help please

    find the interval e^x (1+e^x)^2 dx
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  2. #2
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    Quote Originally Posted by kdstalnaker
    find the interval e^x (1+e^x)^2 dx
    What is the derivative of:

    \frac{(1+e^x)^3}{3}\ ?

    RonL
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  3. #3
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    Quote Originally Posted by kdstalnaker
    find the interval e^x (1+e^x)^2 dx
    I understand it as,
    \int e^x(1+e^x)^2dx.
    Okay, call u=1+e^x, then by substitution,
    \int u^2du=\frac{1}{3}u^3+C
    Thus,
    \frac{1}{3}e^{3x}+C
    Q.E.D.
    Last edited by ThePerfectHacker; February 7th 2006 at 03:07 PM.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    I understand it as,
    \int e^x(1+e^x)^2dx.
    Okay, call u=1+e^x, then by substitution,
    \int u^2du=\frac{1}{3}u^3+C
    Thus,
    \frac{1}{3}e^{3x}+C
    Q.E.D.
    Try differentiating this, then compare it with the integrand. They are
    not the same. Then look at my posting; its a hint. Differentiate what
    I suggest and you will recover the integrand. So what does that tell
    you about the integral?

    RonL

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack
    Try differentiating this, then compare it with the integrand. They are
    not the same. Then look at my posting; its a hint. Differentiate what
    I suggest and you will recover the integrand. So what does that tell
    you about the integral?

    RonL

    RonL
    Not sure where you are getting at?
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  6. #6
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    Quote Originally Posted by ThePerfectHacker
    Not sure where you are getting at?
    What I am getting at is that

    <br />
\frac{d}{dx}\left(\frac{1}{3}e^{3x}+C \right)=e^{3x} \ne  e^x(1+e^x)^2<br />

    and so \frac{1}{3}e^{3x}+C is not the integral of e^x(1+e^x)^2

    But:

    <br />
\frac{d}{dx}\frac{(1+e^x)^3}{3}\ = e^x(1+e^x)^2<br />

    and so \frac{(1+e^x)^3}{3}+C is the integral of e^x(1+e^x)^2

    RonL
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  7. #7
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    Of course! Look at how I used substitution rule. I subsituted e^x for u instead of 1+e^x.
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