求不定积分∫[（2x+1）/(x*x-2x+2)]dx?

原式=∫[（2x-2+3）/(x^2-2x+2)]dx
=∫[（2x-2）/(x^2-2x+2)]dx+∫[3/(x^2-2x+2)]dx
=∫[1/(x^2-2x+2)]d(x^2-2x+2)+3∫{1/[(x-1)^2+1]}d(x-1)
=ln(x^2-2x+2)+3arctan(x-1)+C
楼主所说的∫[（2x-2）/(x^2-2x+2)]dx 到∫[1/(x^2-2x+2)]d(x^2-2x+2）
其实就是典型的凑微分方法 因为(2x-2)dx=d(x^2-2x)=d(x^2-2x+2)
这种很明显要用凑微分的方法嘛