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Math Help - Integral of [e^(x^(1/b))]dx

  1. #1
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    Integral of [e^(x^(1/b))]dx

    I never learned how to integrate this type of function. Can anyone offer insight into how to integrate functions of the form

    e^(x^(1/b)) with respect to x

    and similar functions?

    Thanks
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  2. #2
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    I do not believe this has a closed form. That is probably why you are having trouble.
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  3. #3
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    What do you mean by closed form?

    Also, maybe I should be more specific.

    How do you integrate e^(x^(1/4)) for example?
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  4. #4
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    <br />
\int {e^{x^{\frac{1}{b}} } } dx = \int {e^{\frac{x}{b}} } dx = be^{\frac{x}{b}} + c<br />
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  5. #5
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    Quote Originally Posted by Bastian View Post
    <br />
\int {e^{x^{\frac{1}{b}} } } dx = \int {e^{\frac{x}{b}} } dx = be^{\frac{x}{b}} + c<br />
    Sorry but this 'solution' is complete nonsense. x^{1/b} \neq \frac{x}{b}.
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  6. #6
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    Quote Originally Posted by imabookie3 View Post
    What do you mean by closed form?

    Also, maybe I should be more specific.

    How do you integrate e^(x^(1/4)) for example?
    I don't have time now but it can be done (as can your general case in post #1). See attachment.
    Attached Thumbnails Attached Thumbnails Integral of [e^(x^(1/b))]dx-msp22496262765042408045_44.gif  
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  7. #7
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    Oh, I see, I misread. A general form may be tricky, but for \int e^{x^{\frac{1}{4}}} dx

    We can use parts.

    Let u=4x^{\frac{3}{4}}, \;\ v=e^{x^{\frac{1}{4}}}, \;\ du=3x^{\frac{-1}{4}}dx, \;\ dv=\frac{1}{4}e^{x^{\frac{1}{4}}}x^{\frac{-3}{4}}dx

    Then we can put it together and get:

    4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-3\int\frac{e^{x^{\frac{1}{4}}}}{x^{\frac{1}{4}}}dx

    Now, try applying parts again. Can you do that?.

    You are shooting for this when you're done:

    4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-12x^{\frac{1}{2}}e^{x^{\frac{1}{4}}}+24x^{\frac{1}  {4}}e^{x^{\frac{1}{4}}}-24e^{x^{\frac{1}{4}}}.

    You can factor out the e term to tidy up.

    There is a pattern to these which may lead us to a general form.

    \int e^{x^{\frac{1}{4}}}dx=4e^{x^{\frac{1}{4}}}\left(x^  {\frac{3}{4}}-3x^{\frac{2}{4}}+6x^{\frac{1}{4}}-6\right)

    \int e^{x^{\frac{1}{5}}}dx=5e^{x^{\frac{1}{5}}}\left(x^  {\frac{4}{5}}-4x^{\frac{3}{5}}+12x^{\frac{2}{5}}-24x^{\frac{1}{5}}+24\right)

    \int e^{x^{\frac{1}{6}}}dx=6e^{x^{\frac{1}{6}}}\left(x^  {\frac{5}{6}}-5x^{\frac{4}{6}}+20x^{\frac{3}{6}}-60x^{\frac{2}{6}}+120x^{\frac{1}{6}}-120\right)

    See a pattern you can derive a general form from?. I think so.

    For instance, one way could be be^{x^{\frac{1}{b}}}\left(x^{\frac{b-1}{b}}+(b-1)x^{\frac{b-2}{b}}-(b-2)(b-1)x^{\frac{b-3}{b}}+(b-3)(b-2)(b-1)x^{\frac{b-4}{b}}-....\right)

    There is a start for you.
    Last edited by galactus; October 10th 2008 at 06:48 PM.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Oh, I see, I misread. A general form may be tricky, but for \int e^{x^{\frac{1}{4}}} dx

    We can use parts.

    Let u=4x^{\frac{3}{4}}, \;\ v=e^{x^{\frac{1}{4}}}, \;\ du=3x^{\frac{-1}{4}}dx, \;\ dv=\frac{1}{4}e^{x^{\frac{1}{4}}}x^{\frac{-3}{4}}dx

    Then we can put it together and get:

    4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-3\int\frac{e^{x^{\frac{1}{4}}}}{x^{\frac{1}{4}}}dx

    Now, try applying parts again. Can you do that?.

    You are shooting for this when you're done:

    4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-12x^{\frac{1}{2}}e^{x^{\frac{1}{4}}}+24x^{\frac{1}  {4}}e^{x^{\frac{1}{4}}}-24e^{x^{\frac{1}{4}}}.

    You can factor out the e term to tidy up.

    There is a pattern to these which may lead us to a general form.

    \int e^{x^{\frac{1}{4}}}dx=4e^{x^{\frac{1}{4}}}\left(x^  {\frac{3}{4}}-3x^{\frac{2}{4}}+6x^{\frac{1}{4}}-6\right)

    \int e^{x^{\frac{1}{5}}}dx=5e^{x^{\frac{1}{5}}}\left(x^  {\frac{4}{5}}-4x^{\frac{3}{5}}+12x^{\frac{2}{5}}-24x^{\frac{1}{5}}+24\right)

    \int e^{x^{\frac{1}{6}}}dx=6e^{x^{\frac{1}{6}}}\left(x^  {\frac{5}{6}}-5x^{\frac{4}{6}}+20x^{\frac{3}{6}}-60x^{\frac{2}{6}}+120x^{\frac{1}{6}}-120\right)

    See a pattern you can derive a general form from?. I think so.

    For instance, one way could be be^{x^{\frac{1}{b}}}\left(x^{\frac{b-1}{b}}+(b-1)x^{\frac{b-2}{b}}-(b-2)(b-1)x^{\frac{b-3}{b}}+(b-3)(b-2)(b-1)x^{\frac{b-4}{b}}-....\right)

    There is a start for you.
    Maybe its been too long but

    Let \varphi=x^{\frac{1}{4}}\Rightarrow{\varphi^4=x}

    \therefore{4\varphi^3{d\varphi}=dx}

    \int{e^{x^{\frac{1}{4}}}dx\overbrace{{\mapsto}}^{\  varphi=x^\frac{1}{4}}}\int{e^{\varphi}\cdot{4\varp  hi^3}d\varphi}
    An easy problem?

    and then we can see from that

    \int{e^{x^{\frac{1}{b}}dx}\overbrace{\mapsto}^{\va  rphi=x^{\frac{1}{b}}}\int{e^{\varphi}}\cdot{b\varp  hi^{b-1}}d\varphi}\quad\forall{b}\in\mathbb{Z^+}
    Last edited by Mathstud28; October 10th 2008 at 07:30 PM.
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    Maybe its been too long but

    Let \varphi=x^{\frac{1}{4}}\Rightarrow{\varphi^4=x}

    \therefore{4\varphi^3{d\varphi}=dx}

    \int{e^{x^{\frac{1}{4}}}dx\overbrace{{\mapsto}}^{\  varphi=x^\frac{1}{4}}}\int{e^{\varphi}\cdot{4\varp  hi^3}d\varphi}
    An easy problem?

    and then we can see from that

    \int{e^{x^{\frac{1}{b}}dx}\overbrace{\mapsto}^{\va  rphi=x^{\frac{1}{b}}}\int{e^{\varphi}}\cdot{b\varp  hi^{b-1}}d\varphi}\quad\forall{b}\in\mathbb{Z^+}
    Now I have time, I come back to say these very things and see I'm beaten to the punch. *Socko*
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    Now I have time, I come back to say these very things and see I'm beaten to the punch. *Socko*
    I feel the love
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  11. #11
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    Wow, look who's back from his long hiatus.

    I am referring to you, Mathstud. Where ya' been?.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Wow, look who's back from his long hiatus.

    I am referring to you, Mathstud. Where ya' been?.
    School, MMA, Girls
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  13. #13
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    Quote Originally Posted by mr fantastic View Post
    Sorry but this 'solution' is complete nonsense. x^{1/b} \neq \frac{x}{b}.
    ohh you're right, lapsus
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