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x^2-x+1 = (x-1/2)^2 +3/4
let
x-1/2 = (√3/2)tanu
dx = (√3/2)(secu)^2 du
∫ (x-2)/(x^2-x+1) dx
=(1/2)∫ (2x-1)/(x^2-x+1) dx -(3/2)∫ dx/(x^2-x+1)
=(1/2)ln|x^2-x+1| -(3/2)∫ dx/(x^2-x+1)
=(1/2)ln|x^2-x+1| -(3/2)∫ (√3/2)(secu)^2 du/ [(3/4)(secu)^2]
=(1/2)ln|x^2-x+1| -√3∫du
=(1/2)ln|x^2-x+1| -√3u +C
=(1/2)ln|x^2-x+1| -√3arctan[(2x-1)/√3] +C
let
x-1/2 = (√3/2)tanu
dx = (√3/2)(secu)^2 du
∫ (x-2)/(x^2-x+1) dx
=(1/2)∫ (2x-1)/(x^2-x+1) dx -(3/2)∫ dx/(x^2-x+1)
=(1/2)ln|x^2-x+1| -(3/2)∫ dx/(x^2-x+1)
=(1/2)ln|x^2-x+1| -(3/2)∫ (√3/2)(secu)^2 du/ [(3/4)(secu)^2]
=(1/2)ln|x^2-x+1| -√3∫du
=(1/2)ln|x^2-x+1| -√3u +C
=(1/2)ln|x^2-x+1| -√3arctan[(2x-1)/√3] +C
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