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Math Help - integration

  1. #1
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    integration

    \int\frac{x}{(2x+1)^4} dx
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  2. #2
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    Quote Originally Posted by Punch View Post
    \int\frac{x}{(2x+1)^4} dx
    Have you thought of some substitution?
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  3. #3
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    Quote Originally Posted by Punch View Post
    \int\frac{x}{(2x+1)^4} dx
     \displaystyle \int \frac {x}{(2x+1)^4} \; dx =\frac {1}{2}  \int \frac {1}{(2x+1)^3} \; dx -\frac {1}{2}  \int \frac {1}{(2x+1)^4} \; dx

    use substitution  u = 2x+1 so  du=2dx and so on
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  4. #4
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    sorry, but i dont integrate with substituting. I learnt another method that is... let me give an example

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

    i add the power by 1 and divide the whole function by the new power, and divide it once again by the derivative.

    however, this function has x in both the denominator and the numerator, which is what makes it tough for me
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  5. #5
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    Quote Originally Posted by Punch View Post
    sorry, but i dont integrate with substituting. I learnt another method that is... let me give an example

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

    i add the power by 1 and divide the whole function by the new power, and divide it once again by the derivative.

    however, this function has x in both the denominator and the numerator, which is what makes it tough for me

    solve this than :

     \displaystyle \frac {1}{2} \int \frac {1}{(2x+1)^3} \; dx -\frac {1}{2} \int \frac {1}{(2x+1)^4} \; dx
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  6. #6
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    Quote Originally Posted by yeKciM View Post
    solve this than :

     \displaystyle \frac {1}{2} \int \frac {1}{(2x+1)^3} \; dx -\frac {1}{2} \int \frac {1}{(2x+1)^4} \; dx
    i wouldnt mind solving this but i would want to know how u got this
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  7. #7
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    partial fractions
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  8. #8
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    it isnt possible that i would need to do partial fractions as i have not learnt how to turn such an equation into partial fractions
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  9. #9
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    No, there are no "partial fractions" involved in these problem. In \int\frac{x}{(2x+1)^4} \, dx, \int \frac{1}{2x+1}^3 \, dx, and \int \frac{1}{(2x+1)^4} \, dx, let u= 2x+ 1.
    Last edited by mr fantastic; August 28th 2010 at 10:25 PM. Reason: Fixed some latex, added some dx's.
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  10. #10
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    i am confused
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  11. #11
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    Let u = 2x + 1

    Then x = \frac{u - 1}{2} and dx = du/2.

    So the given problem reduces to

    \frac{1}{2}(\frac{u-1}{u^4})(\frac{du}{2})

    \frac{1}{2}(\frac{u}{u^4}-\frac{1}{u^4})(\frac{du}{2})

    Now find the integration by your method and then substitute the value of u to get the result.
    Last edited by sa-ri-ga-ma; August 28th 2010 at 09:53 PM.
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  12. #12
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    a little question... \int\frac{x}{(2x+1)^4} dx

    can i integrate the above by first taking x out and integrate x\int\frac{1}{(2x+1)^4} dx?
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  13. #13
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    Quote Originally Posted by Punch View Post
    a little question... \int\frac{x}{(2x+1)^4} dx

    can i integrate the above by first taking x out and integrate x\int\frac{1}{(2x+1)^4} dx?
    Yes. You can do it. But you must know the integration by part. Search the text book for the integration by part.
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  14. #14
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    Okay, i have no problem integrating x\int\frac{1}{(2x+1)^4} dx... but my doubt is that in such a way, x remains unintegrated!!!
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  15. #15
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    Quote Originally Posted by Punch View Post
    a little question... \int\frac{x}{(2x+1)^4} dx

    can i integrate the above by first taking x out and integrate x\int\frac{1}{(2x+1)^4} dx?
    NO! You cannot!

    You have been shown how to do this question: The integrand can be written as \displaystyle \frac{1}{2} \left( \frac{1}{(2x + 1)^3} - \frac{1}{(2x + 1)^4} \right) (see below) and you claim in an earlier post to be able to integrate expressions like these. So there shouldn't be any problem.


    \displaystyle \frac{x}{(2x + 1)^4} = \frac{1}{2} \cdot \frac{2x}{(2x + 1)^4} = \frac{1}{2} \cdot \frac{(2x + 1) - 1}{(2x + 1)^4}

    \displaystyle = \frac{1}{2} \left( \frac{2x + 1}{(2x + 1)^4} - \frac{1}{(2x + 1)^4} \right) = \frac{1}{2} \left( \frac{1}{(2x + 1)^3} - \frac{1}{(2x + 1)^4} \right)


    Quote Originally Posted by sa-ri-ga-ma View Post
    Yes. You can do it. But you must know the integration by part. Search the text book for the integration by part.
    This is misleading at best and wrong at worst.
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