∫1/(x^2＋x＋1)dx
=∫1/[(x+1/2)^2+3/4]dx
=4/3*∫1/[(2√3x/3+√3/3)^2+1]*√3/2d(2√3x/3+√3/3)
=8√3/9*∫1/[(2√3x/3+√3/3)^2+1]d(2√3x/3+√3/3)
=8√3/9*arctan(2√3x/3+√3/3)
所以定积分结果为
8√3/9*arctan√3-8√3/9*arctan√3/3
=8√3/9*(π/3-π/6)
=4√3/27*π
希望对楼主有所帮助,