Question

求一个不定积分的题目，谢谢

用凑微分来做么，还是换元法，这种题型的题目，求解

Answer 1

∫(x-3)/(x^2-x+1) dx
=(1/2)∫ (2x-1)/(x^2-x+1) dx - (5/2)∫ dx/(x^2-x+1)
=(1/2)ln|x^2-x+1| - (5/2)∫ dx/(x^2-x+1)
consider
x^2-x+1 = (x-1/2)^2 + 3/4
let
x-1/2 = (√3/2)tany

dx =(√3/2) (secy)^2 dy

∫ dx/(x^2-x+1)
=(2√3/3)∫dy
=(2√3/3)y + C'
=(2√3/3)arctan[(2x-1)/√3]+C'

∫(x-3)/(x^2-x+1) dx
=(1/2)ln|x^2-x+1| - (5/2)∫ dx/(x^2-x+1)
=(1/2)ln|x^2-x+1| - (5√3/3)arctan[(2x-1)/√3]+C