求不定积分∫[（x-1）/（x^2-x+1)]/dx此题属于简单有理式的不定积分

Question

求不定积分 ∫[（x-1）/（x^2-x+1)]/dx
此题属于简单有理式的不定积分

Answer 1

∫(x-1)dx/(x^2-x+1)
=(1/2)∫d(x^2-x+1)/(x^2-2x+1) +(-1/2)∫dx/[(x-1/2)^2+3/4]
=(1/2)ln|x^2-x+1|+(-1/√3)∫d((2x-1)/√3)/[1+[(2x-1)/√3]^2]
=(1/2)ln|x^2-x+1| +(-1/√3)arctan[(2x-1)/√3] +C