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Thread: Integration by substitution

  1. #1
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    Integration by substitution

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

    $

    In this case what would u be? that the integral would be

    $\displaystyle

    \frac{1}{16} \, \int u^{-2} du

    $

    At first glance i thought it would just be
    $\displaystyle
    u=4x^2-4x
    $

    but obviously thats not it.
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  2. #2
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    Quote Originally Posted by simsima_1 View Post
    $\displaystyle
    \int \frac{2x-1}{(4x^2-4x)^{2}} dx\

    $

    In this case what would u be? that the integral would be
    .
    $\displaystyle (4x^2-4x)^2 = 16(x^2 - x)$.
    Let $\displaystyle u=x^2-x$.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    $\displaystyle (4x^2-4x)^2 = 16(x^2 - x)$.
    Let $\displaystyle u=x^2-x$.
    ....dumb question ... but how did u get that? the 16(x^2-x)??
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  4. #4
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    Quote Originally Posted by simsima_1 View Post
    ....dumb question ... but how did u get that? the 16(x^2-x)??
    $\displaystyle (4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$

    EDIT: Mistake fixed.
    Last edited by ThePerfectHacker; Feb 6th 2008 at 07:13 AM.
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  5. #5
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    $\displaystyle (4x^2-4x)^2=(4(x^2-x))^2$
    if $\displaystyle x^2-x = u$ then $\displaystyle (2x-1)dx=du$ so you have

    $\displaystyle \int \frac{du}{(4u)^2}\ = \frac{1}{16} \int \frac{1}{u^2} du\ = \frac{1}{16} \int u^{-2} du\ = -\frac {1}{16u} + C = -\frac{1}{16(x^2-x)} + C$

    The way he wrote it made it seem that $\displaystyle (4x^2-4x)^2=16(x^2-x)$ <-- NOT TRUE

    Below are the correct calulations

    $\displaystyle (4x^2-4x)^2=(4(x^2-x))^2=16(x^2-x)^2$ we used that middle one for our substitution $\displaystyle (4u)^2 = 16u^2$

    Edit : Mistake fixed
    Last edited by Lopared; Feb 6th 2008 at 08:56 AM.
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    $\displaystyle (4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x) = 16(x^2-x)$
    Isn't it $\displaystyle (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$ ?


    Quote Originally Posted by Lopared View Post
    Below are the correct calulations

    $\displaystyle (4x^2-4x)^2=(4(x^2-x))^2=16(x^4-x^2)$
    $\displaystyle (x^2 - x)^2 \neq x^4 - x^2$.
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  7. #7
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    Quote Originally Posted by wingless View Post
    Isn't it $\displaystyle (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$ ?




    $\displaystyle (x^2 - x)^2 \neq x^4 - x^2$.
    yea your right. my mistake (fixed now)
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