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Math Help - Integration by substitution

  1. #1
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    Integration by substitution

    <br />
\int \frac{2x-1}{(4x^2-4x)^{2}} dx\<br /> <br />

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

    <br /> <br />
\frac{1}{16} \, \int u^{-2}   du<br /> <br />

    At first glance i thought it would just be
    <br />
u=4x^2-4x<br />

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

    In this case what would u be? that the integral would be
    .
    (4x^2-4x)^2 = 16(x^2 - x).
    Let u=x^2-x.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    (4x^2-4x)^2 = 16(x^2 - x).
    Let 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)??
    (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; February 6th 2008 at 08:13 AM.
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  5. #5
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    (4x^2-4x)^2=(4(x^2-x))^2
    if x^2-x = u then (2x-1)dx=du so you have

    \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 (4x^2-4x)^2=16(x^2-x) <-- NOT TRUE

    Below are the correct calulations

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

    Edit : Mistake fixed
    Last edited by Lopared; February 6th 2008 at 09:56 AM.
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    (4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x) = 16(x^2-x)
    Isn't it (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

    (4x^2-4x)^2=(4(x^2-x))^2=16(x^4-x^2)
    (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 (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2 ?




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