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Integration by substitution help Watch

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    Integrate 2x/(2x-1)^2 using u=2x-1


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    (Original post by SDavis123)
    Integrate 2x/(2x-1)^2 using u=2x-1
    And you have tried ... ... ...
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    (Original post by TenOfThem)
    And you have tried ... ... ...
    This is what I did:


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    I suppose its clearer to see if I show the step I skipped

    (Integral) x/(2x-1)^2 . 2dx

    Then substitute


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    (Original post by SDavis123)
    This is what I did:
    Fine until

    \dfrac{u+1}{2u^2}

    Then you did nonsense
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    (Original post by TenOfThem)
    Fine until

    \dfrac{u+1}{u^2}

    Then you did nonsense
    But isn't it (u+1)/2u^2? Or is that the same as what you did brcause the 2 cancels with the du as du=2dx


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    (Original post by SDavis123)
    But isn't it (u+1)/2u^2? Or is that the same as what you did brcause the 2 cancels with the du as du=2dx
    yep

    but then you still did nonsense
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    (Original post by TenOfThem)
    yep

    but then you still did nonsense
    but what's wrong with the canceling part?


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    (Original post by SDavis123)
    but what's wrong with the canceling part?


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    Seriously?

    Check this out

    \dfrac{3}{4} = \dfrac{2+1}{2 \times 2} = \dfrac{1}{2}

    That is what you did
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    (Original post by TenOfThem)
    Seriously?

    Check this out

    \dfrac{3}{4} = \dfrac{2+1}{2 \times 2} = \dfrac{1}{2}

    That is what you did
    Argh I have no idea what to do


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    (Original post by SDavis123)
    Argh I have no idea what to do
    \dfrac{u+1}{2u^2} = \dfrac{u}{2u^2} + \dfrac{1}{2u^2}

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    (Original post by TenOfThem)
    \dfrac{u+1}{2u^2} = \dfrac{u}{2u^2} + \dfrac{1}{2u^2}

    Wow I feel thick

    Thanks for your help though


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    (Original post by SDavis123)
    Wow I feel thick

    Thanks for your help though
    np

    sadly it is an error that I see time and time again
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    (Original post by TenOfThem)
    np

    sadly it is an error that I see time and time again
    Just another question, am I correct up to here? I'm unsure if I might have made the same mistake


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    you have multiplied by 2x rather than dividing by 2x when substituting for dx
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    (Original post by TenOfThem)
    you have multiplied by 2x rather than dividing by 2x when substituting for dx
    I'm not sure, my thinking was:

    If I multiply by 2x I obtain 2x^4
    I know du=2xdx
    Therefore all I have to do is find x^3 in terms of u which would be (u-3/4)^3/2

    Then I obtain an integral in terms of u:
    (Integral) ((u-3/4)^3/2)/u^1/2 du


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    (Original post by SDavis123)
    I'm not sure, my thinking was:

    If I multiply by 2x I obtain 2x^4
    I know du=2xdx
    Therefore all I have to do is find x^3 in terms of u which would be (u-3/4)^3/2

    Then I obtain an integral in terms of u:
    (Integral) ((u-3/4)^3/2)/u^1/2 du


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    I do not really understand any of that

    dx = \dfrac{du}{2x}
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    (Original post by SDavis123)
    I'm not sure, my thinking was:

    If I multiply by 2x I obtain 2x^4
    I know du=2xdx
    Therefore all I have to do is find x^3 in terms of u which would be (u-3/4)^3/2

    Then I obtain an integral in terms of u:
    (Integral) ((u-3/4)^3/2)/u^1/2 du


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    dx=du/2x. You cant decide to multiply by 2x Thus its x^3/2x say. Which is 1/2(x^2). So take 1/2 out of the integral, leaving you with \dfrac {1} {2}\int \dfrac {U-4} {\sqrt {U}}

    Thank god
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    This is a pretty similar question, except here I made dx the subject and just substituted, is that ok to do?



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    (Original post by SDavis123)
    This is a pretty similar question, except here I made dx the subject and just substituted, is that ok to do?
    TBH I do not know what you were doing previously

    You want dx as the subject as that is what you are trying to replace
 
 
 
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