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

Answer 2

I= ∫[（x-1）/(x^2-x+1)]/dx
=∫ (x-1)/[(x-1/2)^2+ 3/4]/dx
let
x-1/2 = (√3/2)tana
dx = (√3/2)(seca)^2 da
I=∫ (x-1)/[(x-1/2)^2+ 3/4]/dx
= (4/3)∫ { [(√3/2)tana- 1/2] /(seca)^2} (√3/2)(seca)^2 da
= 2√3/3∫[(√3/2)tana- 1/2] da
=2√3/3 [-(√3/2)ln|cosa| - a/2] +C

where a= arctan(2√3/3 (x-1/2))