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Math Help - problem in integration

  1. #1
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    problem in integration

    integrate (x-1) / (2x+1)^3 dx

    i found that it will be  -1/2 (2x+1)^-^1 + 3/4 (2x+1)^-^2 +c

    but i have guide answer :  -1/4 (2x+1)^-^1 + 3/8 (2x+1)^-^2 +c
    Last edited by mido22; February 20th 2013 at 04:25 AM.
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  2. #2
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    Re: problem in integration

    split the fraction so ( x/(2x+1) -1/(2x + 1))dx
    now let u= 2x + 1
    x =(u -1)/2
    integrate separately using the above substitution on the first piece
    the second is just -1/u integrated
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  3. #3
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    Re: problem in integration

    Uhm...

    \int \frac{x-1}{2x+1}\, dx=\int \frac{1}{2}\cdot \frac{2(x-1)}{2x+1}\, dx=

    =\frac{1}{2}\int \frac{2x+1-3}{2x+1}\, dx=\frac{1}{2}\int \left (1-\frac{3}{2x+1}  \right )\, dx=

    =\frac{1}{2}\left (\int 1\, dx -\int\frac{3}{2} \cdot \frac{2}{2x+1}  \right )\, dx  \right )=

    =\frac{1}{2}\left ( x- \frac{3}{2} \ln (2x+1)\right )+C
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  4. #4
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    Re: problem in integration

    sorry all but there is power 3 on (2x+1) in the problem
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  5. #5
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    Re: problem in integration

    That doesnt make much of a difference, use veileens method with the power of 3
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  6. #6
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    Re: problem in integration

    Quote Originally Posted by MathJack View Post
    That doesnt make much of a difference, use veileens method with the power of 3
    but i can't understand the last two steps in his solution how 2/(2x+1) become (2x+1) ???
    sorry but this is my first day to learn about integration ....
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  7. #7
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    Re: problem in integration

    either my solution correct or the guide answer is correct??
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  8. #8
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    Re: problem in integration

    but i can't understand the last two steps in his solution how 2/(2x+1) become (2x+1) ???
    *her ^^'
    Do you know the integration/derivative rules? If not, you should first learn them.


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

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

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

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

    =-\frac{1}{4}\cdot \frac{1}{2x+1}-\frac{3}{8}\cdot \frac{1}{(2x+1)^2}+C

    Their answer is the correct one. If that's all you wanted to know, you could have used WolframAlpha.
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  9. #9
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    Re: problem in integration

    Mido22

    expand the function to get 1/2(2x+1)^2 -1/2(2x+1)^3 and then integrate to get (1-4x)/8(2x+1)^2 +C
    it is easy.
    MINOAS
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