integrate 2/((2x-7)^2) from x=4 to 6

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  1. Let 2x -7 = u
    so dx = du/2
    Integrand becomes
    2/u^2*(du/2) = du/u^2 from u = 1 to 5.
    = -1/u @ u = 5 -
    (-1/u @ u = 1)
    = 1 - 1/5 = 4/5

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  2. 2/((2x-7)^2)
    = 2(2x-7)^-2

    integral of that is
    -(2x-7)^-1 or -1/(2x-7)

    and its value form x=4 to 6
    = -1/5 + 1
    = 4/5

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