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Math Help - Another Integration Problem

  1. #1
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    Another Integration Problem

    \int(x^{2} + 2x - 1)^{3}2x dx

    du = 2x + 2

    Now putting x to the 4th power, and 4 in the denominator.

    \int \frac{(x^{2} + 2x - 1)^{4}}{4} + C Not the Final Answer

    However, that's the not the final answer. Some special manipulation needs to be done. That's because the du is larger than dx (of the original problem). It think some sort of division needs to be tried.

    Let's try it this way:

    \frac{(x^{2} + 2x - 1)}{1/2} * \frac{ (2x + 2)}{2}
    Last edited by Jason76; November 21st 2012 at 08:35 PM.
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  2. #2
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    Re: Another Integration Problem

    I would expand and integrate term by term.
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    Re: Another Integration Problem

    Quote Originally Posted by MarkFL2 View Post
    I would expand and integrate term by term.
    How? Give me a head start. But don't do the whole thing.

    Note: The du almost matches the dx (in the original problem). However, there is an extra + 2 that needs to be gotten rid of.

    If the dx was 4x dx then division by 2 would be the key. But in this problem, that isn't the case.

    Ok, let's try to get rid of the + 2 by subtraction.

    [(x^{2} + 2x - 1) - \frac{1}{2}] [(2x + 2) - 2]
    Last edited by Jason76; November 21st 2012 at 08:47 PM.
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    MHF Contributor MarkFL's Avatar
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    Re: Another Integration Problem

    (x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1

    Now, distribute the x onto that, pull the two out front, then integrate term by term.
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    Re: Another Integration Problem

    Quote Originally Posted by MarkFL2 View Post
    (x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1

    Now, distribute the 2x onto that, pull the two out front, then integrate term by term.
    \int(x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1

    \int 2x(x^6+6x^5+9x^4-4x^3-9x^2+6x-1)

    \int 2x^{7} + 12x^{6} + 9x^{5} - 8x^{4} - 18x^{3} + 12x^{2} - 2x

    Integrating term by term:

    \int \frac{2x^{}8}{8} + \frac{12x^{7}}{7} + \frac{9x^{6}}{6} - \frac{8x^{5}}{5} - \frac{18x^{4}}{4} + \frac{12x^{3}}{3} - \frac{2x^{2}}{2}

    Now what?
    Last edited by Jason76; November 21st 2012 at 09:04 PM.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Another Integration Problem

    I would choose to leave the 2 out front as a factor, but either way will work.
    Last edited by MarkFL; November 21st 2012 at 09:07 PM.
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    Re: Another Integration Problem

    What's the next step? I did what you said. Is that how it's supposed to look so far?
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  8. #8
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    Re: Another Integration Problem

    Now, simply integrate term by term using the power rule. I see you have edited your post...once you have integrated, you don't want to use the integration symbol, that is removed when you write the anti-derivative. So, remove the integration symbol, reduce any fractions and add the constant of integration.
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    Re: Another Integration Problem

    Reduced Fractions and added + C

     \frac{x^{}8}{4} + \frac{12x^{7}}{7} + \frac{3x^{6}}{2} - \frac{8x^{5}}{5} - \frac{9x^{4}}{2} + \frac{4x^{3}}{1} - \frac{x^{2}}{1} + C

    But when do we use this method of integration as opposed to the other ways?
    Last edited by Jason76; November 21st 2012 at 09:25 PM.
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Another Integration Problem

    Where the denominators are 1, I would omit them.

    I looked at methods of substitution, but did not readily find a method that would work. There may be a way, but my inclination was to just expand and integrate.
    Last edited by MarkFL; November 21st 2012 at 09:58 PM.
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