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Math Help - Stuck on Integration by Parts

  1. #1
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    Stuck on Integration by Parts

    Question 1
    \int e^\frac1x x^2
    First thing I did was attempt integration by parts
    u=\frac 1x \ du=\frac 1{x^2} \ dv=x^2 \ v=\frac {x^3}3

    Using the formula
    UV-\int VdU

    I come up with
    (\frac 1x) (\frac13 x^3) - \frac13 \int \frac{x^3}{x^2}

    Here is where I get stuck with the integration, I think I can move the
    x^2
    up top and make it
    x^-2

    So I am left with
    \int {x^3}{x^-2}
    Above is Supposed to be a negative 2, but couldnt find the code for it.

    So if I use u sub
    U=x^3 \ dU=3x^2

    The 3 I can handle but not sure what to do about the fact that my exponent is negative, and my dU has a positive.

    Any help?

    Thanks
    Rich
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  2. #2
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    Quote Originally Posted by Spoolx View Post
    Question 1
    \int e^\frac1x x^2
    First thing I did was attempt integration by parts
    u=\frac 1x \ du=\frac 1{x^2} \ dv=x^2 \ v=\frac {x^3}3

    Using the formula
    UV-\int VdU

    I come up with
    (\frac 1x) (\frac13 x^3) - \frac13 \int \frac{x^3}{x^2}

    Here is where I get stuck with the integration, I think I can move the
    x^2
    up top and make it
    x^-2

    So I am left with
    \int {x^3}{x^-2}
    Above is Supposed to be a negative 2, but couldnt find the code for it.

    So if I use u sub
    U=x^3 \ dU=3x^2

    The 3 I can handle but not sure what to do about the fact that my exponent is negative, and my dU has a positive.

    Any help?

    Thanks
    Rich
    Dear Spoolx,

    This integration could not be found using elementary functions. See Wolfram Mathematica Online Integrator
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  3. #3
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    Thanks for the reply, I will do some more reading in the book, however its in the integration by parts section, so I figured it could be done that way.

    How would you recommend solving this problem if I was given it on an exam?
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  4. #4
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    Quote Originally Posted by Spoolx View Post
    Thanks for the reply, I will do some more reading in the book, however its in the integration by parts section, so I figured it could be done that way.

    How would you recommend solving this problem if I was given it on an exam?
    I think there typo in the book. Integration by parts won't work. The maximum you could progress through integration by parts method is,

    \displaystyle\int x^2 e^{\frac{1}{x}}dx=\frac{x^3}{3}e^{\frac{1}{x}}+\fr  ac{1}{6}x^2 e^{\frac{1}{x}}+\frac{1}{6}\int e^{\frac{1}{x}}dx

    Then to evaluate, \int e^{\frac{1}{x}}dx you will need to know about the "Exponential integral". So I hope this will not be given in a exam at your level.
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  5. #5
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    But... parts once more takes you to

    \frac{1}{6} \int \frac{1}{x} e^{\frac{1}{x}} dx

    which is the ExpIntegralEi - Wolfram Mathematica 8 Documentation the exponential integral of 1/x.

    http://www.wolframalpha.com/input/?i...281%2Fx%29%2Fx

    So you can use parts to arrive at Wolfram's result, even though the special function is switching off the 'show steps' option that Wolfram usually offers.

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,



    ... is lazy integration by parts, doing without u and v.


    (But, oh dear, the new MHF page layout isn't friendly to wide diagrams.)

    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; January 1st 2011 at 05:36 AM.
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  6. #6
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    Quote Originally Posted by tom@ballooncalculus View Post
    But... parts once more takes you to

    \frac{1}{6} \int \frac{1}{x} e^{\frac{1}{x}} dx

    which is the ExpIntegralEi - Wolfram Mathematica 8 Documentation

    So you can use parts to arrive at Wolfram's result, even though the special function is switching off the 'show steps' option that Wolfram usually offers.
    Hey! Thats a suprise. The exponential integral is defined as Ei(x)=\displaystyle\int{\frac{e^{\frac{1}{x}}}{x}}  dx. Therefore you are correct. The answer could be obtained easily. And thank you very much for pointing that, I learnt something new.

    To Spoolx,

    Sorry if I confused you in my previous post. When I saw the Ei abbreviation in wolfram thought it was something which we cant obtain. The problem is correct.
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  7. #7
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    Quote Originally Posted by Spoolx View Post
    Question 1
    \int e^\frac1x x^2
    First thing I did was attempt integration by parts
    u=\frac 1x \ du=\frac 1{x^2} \ dv=x^2 \ v=\frac {x^3}3
    Well, this is incorrect. You use 'integration by parts' to integrate \int udv which, with this u and dv, would be \int \frac{x^2}{x}dx= \int x dx. What you have is \int e^u dv and the integration by parts formula does not apply.

    Using the formula
    UV-\int VdU

    I come up with
    (\frac 1x) (\frac13 x^3) - \frac13 \int \frac{x^3}{x^2}

    Here is where I get stuck with the integration, I think I can move the
    x^2
    up top and make it
    x^-2

    So I am left with
    \int {x^3}{x^-2}
    Above is Supposed to be a negative 2, but couldnt find the code for it.

    So if I use u sub
    U=x^3 \ dU=3x^2

    The 3 I can handle but not sure what to do about the fact that my exponent is negative, and my dU has a positive.

    Any help?

    Thanks
    Rich
    Follow Math Help Forum on Facebook and Google+

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