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    integrate:
    x/(x-2)^2

    answer: ln(x-1) - x/(x-1)

    I got: x/x-1 - ln(x-1)

    step by step method so I can see where I went wrong would be great and i'll give +ve rep
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    Integration by substitution. Know how to do that?

    \int\dfrac{x}{(x-2)^2}dx

    Let u = x-2
    \frac{du}{dx} = 1
    \frac{dx}{du} = 1

    x = u + 2

    \int\dfrac{u+2}{u^2}\times 1 du

    \int(u+2)u^{-2}du

    \int(u^{-1}+2u^{-2})du

    = lnu - 2u^{-1}

    = lnu - \frac{2}{u}

    = ln(x-2) - \frac{2}{x-2}

    AhH!. oops. that's my answer. latex confuses me
    But I'm sure that's right, after looking voer it again
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    (Original post by maxfire)
    Integration by substitution. Know how to do that?

    \int\dfrac{x}{(x-2)^2}dx

    Let u = x-2
    \frac{du}{dx} = 1
    \frac{dx}{du} = 1

    x = u + 2

    \int\dfrac{u+2}{u^2}\times 1 du

    \int(u+2)u^{-2}du

    \int(u^{-1}+2u^{-2})du

    = lnu - 2u^{-1}

    = lnu - \frac{2}{u}

    = ln(x-2) - \frac{2}{x-2}

    AhH!. oops. that's my answer. latex confuses me
    But I'm sure that's right, after looking voer it again
    That's definately not right...
    I wouldn't do subsitiution as in the question it says use integration by parts
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    My method

    u = x
    du/dx = 1

    dv/dx = 1/(x-2)^2 =(x-1)^-2
    v=-1/x-1

    answer = -x/(x-1) - integral of 1/x-2
    =-x(x-1) + ln(x-1)

    WOO i did it >.>
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    (Original post by SmileyGurl13)
    My method

    u = x
    du/dx = 1

    dv/dx = 1/(x-2)^2 =(x-1)^-2
    v=-1/x-1

    answer = -x/(x-1) - integral of 1/x-2
    =-x(x-1) + ln(x-1)

    WOO i did it >.>
    Either I'm going mental, or there's something not quite right here.
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    Let  u = x \implies \dfrac{\mathrm{d}u}{\mathrm{d}x} = 1 and  \dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{1}{(x-2)^2} \implies v = \dfrac{-1}{x-2}

     \displaystyle\int \dfrac{x}{(x-2)^2} \ \mathrm{d}x = \displaystyle\int u\dfrac{\mathrm{d}v}{\mathrm{d}x  } \ \mathrm{d}x = uv - \displaystyle\int v\dfrac{\mathrm{d}u}{\mathrm{d}x  } \ \mathrm{d}x

     \iff \displaystyle\int \dfrac{x}{(x-2)^2} = x \times \dfrac{-1}{x-2} - \displaystyle\int \dfrac{-1}{x-2} \times 1 \ \mathrm{d}x

    I'm sure you can finish it from there.
    Spoiler:
    Show
    So, the answer should be:

     \ln |x-2| - \dfrac{x}{x-2}

    So, I disagree with the answer in the book or the answer given by the teacher. :erm:
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    out of interest, what are people typing up the solutions in? is it a maths program, or just word?
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    (Original post by edmhaslam)
    out of interest, what are people typing up the solutions in? is it a maths program, or just word?
    It's an inbuilt program in TSR called LateX. You can learn the basics here and then find some more codes here.
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    (Original post by GHOSH-5)
    Let and





    I'm sure you can finish it from there.
    Ok, but how did change to ? Correct me if I'm wrong, but I'm quite sure that \displaystyle\int\frac1{(x-2)^2} = \frac{-1}{x-2} .
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    (Original post by meatball893)
    Ok, but how did change to ? Correct me if I'm wrong, but I'm quite sure that \displaystyle\int\frac1{(x-2)^2} = \frac{-1}{x-2} .
    Whoops, my eyes are going mad; I'll edit that post! :o:
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    Three observations:

    1. We've all (including me) forgotten our constants! :eek:
    2. The book has a typo. If you replace 1s with 2s and imagine a constant it's all OK!
    3. maxfire's solution was correct (though missing a constant). It is certainly not "definitely not right."
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    \int\frac{x}{(x-2)^2}dx
    sweet
    sorry not helping just wanted to try latex out :P
    it's a bit of a chore though :/
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    If you put \displaystyle before \int it looks a lot nicer.

    You get \displaystyle\int f(x) dx instead of \int f(x) dx.

    And it's not so bad. You just have to get used to it.
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    (Original post by meatball893)
    Either I'm going mental, or there's something not quite right here.
    the dv/dx is gathered by the question as in it's the other part of it and then you have integrate it to work out v
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    (Original post by SmileyGurl13)
    the dv/dx is gathered by the question as in it's the other part of it and then you have integrate it to work out v
    I am familiar with integration by parts. I was merely questioning how 1/[(x-2)^2] became (x-1)^(-2).
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    (Original post by meatball893)
    I am familiar with integration by parts. I was merely questioning how 1/[(x-2)^2] became (x-1)^(-2).
    It's a mistake/typo in the answers you've been given.
 
 
 
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