∫(2x-3)/(x²-x+3) dx
=∫(2x-1)/(x²-x+3) dx-2∫dx/(x²-x+3)
=∫d(x²-x+3)/(x²-x+3)-2∫1/[(x-1/2)²+11/4] d(x-1/2)，可设x-1/2=√(11/4)*tanθ求得
=ln|x²-x+3|-2*1/√(11/4)*arctan[1/√(11/4)*(x-1/2)]+C
=ln|x²-x+3|-4/√11*arctan[2/√11*(x-1/2)]+C
=ln|x²-x+3|-(4/√11)arctan[(2x-1)/√11]+C
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