integral (4x^2-3x+2)/(4x^2-4x+3) dx

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  1. long division shows that we have

    ∫ 1 + (x-1)/(4x^2-4x+3)
    = ∫ 1 + (x-1)/[(2x-1)^2 + 2] dx

    let (2x-1)^2 = 2tan^2θ, so
    (2x-1)^2 + 2 = 2tan^2θ+2 = 2sec^2θ

    2x-1 = √2 tanθ
    x = (√2 tanθ+1)/2
    x-1 = (√2 tanθ-1)/2
    dx = 1/2 √2 sec^2θ

    ∫ 1 + (x-1)/[(2x-1)^2 + 2] dx
    = ∫ (1 + (√2 tanθ-1)/(4sec^2θ)) 1/√2 sec^2θ dθ
    . . .
    = 1/8(x + 1/8 log(4x^2-4x+3) - 1/(4√22) arctan (2x-1)/√2

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    posted by Steve

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