请问图中画红线的式子是怎么得出来的呢，给个详细过程。谢谢
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对上一步后面一个求积分，即：∫[(2-x)/(x²-x+1)]dx
=-∫[(x-2)/(x²-x+1)]dx
=-∫[(1/2)*(2x-1)-(3/2)]/[(x²-x+1)]dx
=-(1/2)∫[(2x-1)/(x²-x+1)]dx+(3/2)∫[1/(x²-x+1)]dx
=(-1/2)∫[1/(x²-x+1)]d(x²-x+1)+(3/2)∫[1/(x²-x+1)]dx
——其实就是“凑微分”！！！
=-∫[(x-2)/(x²-x+1)]dx
=-∫[(1/2)*(2x-1)-(3/2)]/[(x²-x+1)]dx
=-(1/2)∫[(2x-1)/(x²-x+1)]dx+(3/2)∫[1/(x²-x+1)]dx
=(-1/2)∫[1/(x²-x+1)]d(x²-x+1)+(3/2)∫[1/(x²-x+1)]dx
——其实就是“凑微分”！！！
追问
能帮我答答其他的题吗，谢谢
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