高数题目定积分：上限1下限0 1⼀（x^2＋x＋1）dx

∫(0~1) dx/(x² + x + 1)
= ∫(0~1) dx/[(x + 1/2)² + 3/4]
令x + 1/2 = (√3/2)tanz <=> tanz = (2x + 1)/√3，dx = (√3/2)sec²z dz
当x = 0，tanz = 1/√3 => z = π/6
当x = 1，tanz = 3/√3 => z = π/3
原式= ∫(π/6~π/3) (√3/2)sec²z/[(3/4)tan²z + 3/4] dz
= ∫(π/6~π/3) (√3/2)sec²z/[(3/4)sec²z] dz
= (√3/2)(4/3)∫(π/6~π/3) dz
= (2/√3) * z|_(π/6^π/3)
= (2/√3) * [π/3 - π/6]
= π/(3√3) = (π√3)/9
楼主，我答的比楼上更详细，所以应该采纳我。

楼上做的答案是错了，以后不做你的题了！！

∫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，望采纳！