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Thread: Integration of dx/[(1-2x)^2*(3-x)]

  1. #1
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    Integration of dx/[(1-2x)^2*(3-x)]

    I have attempted to solve this integration problem by parts but I have not been able to arrive at the final answer. Can anyone help?

    Problem: integration of dx/[(1-2x)^2*(3-x)]



    Answer: 1/[5*(1-2x)] + (1/25)*Ln[(2x-1)/(x-3)]
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by gnameni View Post
    I have attempted to solve this integration problem by parts but I have not been able to arrive at the final answer. Can anyone help?

    Problem: integration of dx/[(1-2x)^2*(3-x)]



    Answer: 1/[5*(1-2x)] + (1/25)*Ln[(2x-1)/(x-3)]
    I would say you should prefer intigration by parts

    $\displaystyle \int{\frac{dx}{(1-2x)^2 (3-x)}} $

    $\displaystyle = \frac{1}{-4}\int{\frac{dx}{(x-1/2)^2(x-3)}}$

    Now $\displaystyle \frac{1}{(x-1/2)^2(x-3)} = \frac{A}{x-1/2)^2} + \frac{B}{x-3} $

    Find A And B
    and then integrate

    $\displaystyle \frac{1}{-4}[\int{ \frac{A}{(x-1/2)^2}} +\int{ \frac{B}{x-3}}] $
    Last edited by ADARSH; Mar 17th 2009 at 12:22 AM. Reason: typo
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  3. #3
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    I do not think that
    dx/(1-2x)^2 equals -1/4(dx/(x-1/2)^2)
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by gnameni View Post
    I do not think that
    dx/(1-2x)^2 equals -1/4(dx/(x-1/2)^2)
    Neither do I think so (if you are talking about negative sign )
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  5. #5
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    Yes- but that is what you wrote.

    I think that the correct expression for the integration of dx/[(1-2x)^2*(3-x)]

    is -1/2*integral(dx/[(x-1/2)^2*(x-3)])

    How would one go about solving: integral of A/(X-1/2)^2 + integral of B/(x-3)?
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  6. #6
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    Quote Originally Posted by gnameni View Post
    Yes- but that is what you wrote.

    I think that the correct expression for the integration of dx/[(1-2x)^2*(3-x)]

    is -1/2*integral(dx/[(x-1/2)^2*(x-3)])
    If you are trying to say that

    $\displaystyle
    (1-2x)^2(3-x) = -2(x-\frac{1}{2})^2 (x-3)$

    than I am sorry sun will not rise from left,try again
    Twinkle Twinkle Little Star..

    How would one go about solving: integral of A/(X-1/2)^2 + integral of B/(x-3)?
    Before solving question you have asked you must have solved these types of question

    but just for an example I will give A= 1 & B = 1

    $\displaystyle \int{\frac{dx}{(x-1/2)^2}} $

    Put (x-1/2) = t

    dx =dt

    Hence integral becomes

    $\displaystyle \int{\frac{dt}{(t)^2}} $

    $\displaystyle = \frac{-1}{t}+c= \frac{-1}{x-1/2} +c$

    doing the same for second integral

    $\displaystyle \int{\frac{dx}{x-3}} = ln|x-3| +c$
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  7. #7
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    Hllo, gnameni!

    Let's take it from the top . . .


    $\displaystyle \int \frac{dx}{(1-2x)^2(3-x)}$

    $\displaystyle \text{Answer: }\;\frac{1}{5(1-2x)} + \frac{1}{25}\ln\left[\frac{2x-1}{x-3}\right]$
    Partial Fractions . . .

    We have: . $\displaystyle \frac{1}{(1-2x)^2(3-x)} \;=\;\frac{A}{1-2x} + \frac{B}{(1-2x)^2} + \frac{C}{3-x} $

    Then: . $\displaystyle 1 \;=\;A(1-2x)(3-x) + B(3-x) + C(1-2x)^2$


    Let $\displaystyle x = \tfrac{1}{2}\!:\quad 1 \;=\;A(0) + B(\tfrac{5}{2}) + C(0) \quad\Rightarrow\quad \boxed{B = \tfrac{2}{5}}$ .[1]

    Let $\displaystyle x = 3\!:\quad 1 \;=\;A(0) + B(0) + C(25) \quad\Rightarrow\quad\boxed{ C = \tfrac{1}{25}}$ .[2]

    Let $\displaystyle x = 0\!:\quad 1 \;=\;A(3) + B(3) + C(1) \quad\Rightarrow\quad 3A + 3B + C \;=\;1$
    . . Substitute [1] and [2]: .$\displaystyle 3A + \tfrac{6}{5} + \tfrac{1}{25} \:=\:1 \quad\Rightarrow\quad\boxed{ A = \text{-}\tfrac{2}{25}}$


    Hence, we have: .$\displaystyle \frac{\text{-}\frac{2}{25}}{1-2x} + \frac{\frac{2}{5}}{(1-2x)^2} + \frac{\frac{1}{25}}{3-x}$


    Integrate: . $\displaystyle -\frac{2}{25}\int\frac{dx}{1-2x} + \frac{2}{5}\int(1-2x)^{-2}dx + \frac{1}{25}\int\frac{dx}{3-x} $

    . . . . . . . . $\displaystyle = \;\frac{1}{25}\ln(1-2x) + \frac{1}{5}(1-2x)^{-1} -\frac{1}{25}\ln(3-x) + C $

    . . . . . . . . $\displaystyle = \;\frac{1}{5}\!\cdot\!\frac{1}{1-2x} + \frac{1}{25}\bigg[\ln(1-2x) - \ln(3-x)\bigg] + C$

    . . . . . . . . $\displaystyle = \;\frac{1}{5(1-2x)} + \frac{1}{25}\ln\left[\frac{1-2x}{3-x}\right] + C $


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