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December 19, 2014

December 19, 2014

Posted by **Anonymous** on Tuesday, December 4, 2012 at 4:00am.

- Calculus -
**Steve**, Tuesday, December 4, 2012 at 5:00amlong 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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