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Math Help - Integrating an irrational function

  1. #1
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    Integrating an irrational function

    The question

    \int{\frac{u^{14}}{u + 1}}

    My attempt
    Normally I'd use polynomial division to produce a rational integrand, but as you can see, the large power of u causes the division process to become quite lengthy. Is there a more efficient way of producing an easily solvable integrand?

    Thanks.
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  2. #2
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    If you don't want to use long-division, you can just do a trick...

    \displaystyle \frac{u^{14}}{u + 1} = \frac{u^{14} + u^{13} - u^{13}}{u + 1}

    \displaystyle = \frac{u^{13}(u + 1)}{u + 1} - \frac{u^{13}}{u + 1}

    \displaystyle = u^{13} - \frac{u^{13}}{u + 1}

    \displaystyle = u^{13} - \frac{u^{13} + u^{12} - u^{12}}{u + 1}

    \displaystyle = u^{13} - \frac{u^{12}(u + 1)}{u + 1} + \frac{u^{12}}{u + 1}

    \displaystyle = u^{13} - u^{12} + \frac{u^{12}}{u + 1} etc...
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    An alternative:

    Using the substitution t=u+1 we can obtain a closed solution:

    \displaystyle\int \dfrac{u^{14}du}{u+1}=\displaystyle\int \dfrac{(t-1)^{14}dt}{t}=\displaystyle\int \dfrac{1}{t}\sum_{k=0}^{14}{(-1)^{n-k}t^k}dt}=\displaystyle\int \sum_{k=0}^{14}{(-1)^{n-k}t^{k-1}dt}=

    \log |t|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}t^k}{k}+C=\log |u+1|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}(u+1)^k}{k}+C


    Fernando Revilla

    Edited: I forgot the binomial coefficients.
    Last edited by FernandoRevilla; January 8th 2011 at 09:41 AM.
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  4. #4
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    Quote Originally Posted by FernandoRevilla View Post
    An alternative:

    Using the substitution t=u+1 we can obtain a closed solution:

    \displaystyle\int \dfrac{u^{14}du}{u+1}=\displaystyle\int \dfrac{(t-1)^{14}dt}{t}=\displaystyle\int \dfrac{1}{t}\sum_{k=0}^{14}{(-1)^{n-k}t^k}dt}
    This is incorrect. (t-1)^{14} is not equal to \sum (-1)^{n-k}t^k. You forgot the binomial coefficients. (t-1)^{14}= \sum_{k=0}^{14} (-1)^{m-k}\begin{pmatrix}14 \\ k\end{pmatrix} t^k.

    = \displaystyle\int \sum_{k=0}^{14}{(-1)^{n-k}t^{k-1}dt}=

    \log |t|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}t^k}{k}+C=\log |u+1|+\displaystyle\sum_{k=1}^{14}\dfrac{(-1)^{n-k}(u+1)^k}{k}+C


    Fernando Revilla
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    This is incorrect. (t-1)^{14} is not equal to \sum (-1)^{n-k}t^k. You forgot the binomial coefficients. (t-1)^{14}= \sum_{k=0}^{14} (-1)^{m-k}\begin{pmatrix}14 \\ k\end{pmatrix} t^k.
    O ... ps !, of course. More than incorrect. I need a mathematical psychiatrist I see.


    Fernando Revilla
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    I need a mathematical psychiatrist I see.


    Fernando Revilla

    Hahaha...
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Glitch View Post
    The question

    \int{\frac{u^{14}}{u + 1}}

    My attempt
    Normally I'd use polynomial division to produce a rational integrand, but as you can see, the large power of u causes the division process to become quite lengthy. Is there a more efficient way of producing an easily solvable integrand?

    Thanks.
    ... you forgot "du" at your integral...
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