如图所示
=∫(x+1)/(x²+x+1)-x/(x²+1)dx前半部换元u=x+1/2简化=∫u/(u²+3/4)+(1/2)/(u²+3/4)du-(1/2)ln(x²+1)=(1/2)ln(u²+3/4)+(1/√3)arctan(2u/√3)-(1/2)ln(x²+1)+C
