Question

求不定积分 x/(x^2+x+1)^0.5

我用3^0.5/2tanx=x+1/2, 算出来化简下来最终是3^0.5(secx)/2，而答案很复杂，是(x^2+x+1)^0.5-0.5ln((x^2+x+1)^0.5+x+1/2)
求助！多谢

Answer 1

∫ x/√(x² + x + 1) dx
= ∫ x/√[(x + 1/2)² + 3/4] dx
x + 1/2 = √(3/4) tany => dx = √3/2 sec²y dy
原式 = ∫ (√3/2 tany - 1/2)/√[(3/4) tan²y + 3/4] (√3/2 sec²y dy)
= ∫ (√3/2 tany - 1/2)/(√3/2 secy) (√3/2 sec²y) dy
= ∫ (√3/2 tany - 1/2)secy dy
= (√3/2)∫ secy tany dy - (1/2)∫ secy dy
= (√3/2)secy - (1/2)ln|secy + tany| + C
= (√3/2) [2√(x² + x + 1)/√3] - (1/2)ln| [2√(x² + x + 1)]/√3 + (2x + 1)/√3] | + C
= √(x² + x + 1) - (1/2)ln| 2x + 1 + 2√(x² + x + 1) | + C
或 = √(x² + x + 1) - (1/2)ln| x + 1/2 + √(x² + x + 1) | + C
Note：
∵tany = (x + 1/2)/(√3/2) = (2x + 1)/√3
∴siny = (2x + 1)/[2√(x² + x + 1)]
∴cosy = √3/[2√(x² + x + 1)]，secy = [2√(x² + x + 1)]/√3