∫x/(x^2-x+1)dx用凑微分法怎么求?

x/(x^2-x+1) = (x -1/2) /(x^2-x+1) + (1/2) /(x^2-x+1)∫(x -1/2) /(x^2-x+1) dx 凑微分,u = (x^2-x+1)= (1/2)∫du / u = (1/2) lnu + C = (1/2) ln (x^2-x+1) + C∫(1/2) /(x^2-x+1) dx = (1/2)∫dx / [(x-1/2)
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