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Math Help - integral

  1. #1
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    integral

    Thanks!

    I have a similar problem, I am trying to solve the way you suggested:

    <br />
\displaystyle\int \frac{e^{6 x}}{e^{12 x} + 9} dx<br />

    let U=e^6x

    <br />
\displaystyle\int \frac{U}{U^2+9} <br />
    now, if I pull a 1/9 out of the integral

    I can get
    <br />
\displaystyle\frac{1}{9} \int\frac{U}{\frac{U^2}{9}+1}<br />

    Aghh, and there is ALMOST an arctan there, if it wasnt for that U in the numerator!

    What do I do here? Am I not supposed to use this same method?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Vamz View Post
    Thanks!

    I have a similar problem, I am trying to solve the way you suggested:

    <br />
\displaystyle\int \frac{e^{6 x}}{e^{12 x} + 9} dx<br />

    let U=e^6x

    <br />
\displaystyle\int \frac{U}{U^2+9} <br />
    now, if I pull a 1/9 out of the integral

    I can get
    <br />
\displaystyle\frac{1}{9} \int\frac{U}{\frac{U^2}{9}+1}<br />

    Aghh, and there is ALMOST an arctan there, if it wasnt for that U in the numerator!

    What do I do here? Am I not supposed to use this same method?
    e^{6x}=3\tan(\theta)

    I think this should work. I didn't try it.
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  3. #3
    Behold, the power of SARDINES!
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    Yuma, AZ, USA
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    Quote Originally Posted by Vamz View Post
    Thanks!

    I have a similar problem, I am trying to solve the way you suggested:

    <br />
\displaystyle\int \frac{e^{6 x}}{e^{12 x} + 9} dx<br />

    let U=e^6x

    <br />
\displaystyle\int \frac{U}{U^2+9} <br />
    now, if I pull a 1/9 out of the integral

    I can get
    <br />
\displaystyle\frac{1}{9} \int\frac{U}{\frac{U^2}{9}+1}<br />

    Aghh, and there is ALMOST an arctan there, if it wasnt for that U in the numerator!

    What do I do here? Am I not supposed to use this same method?
    It should not be

    if u=e^{6x} \implies du=6e^{6x}dx This gives the integtal

    \displaystyle \frac{1}{6}\int\frac{du}{u^2+9}
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