Question

(x+1)/根号(x^2+x+1)dx,求不定积分

Answer 1

∫ (x + 1)/√(x² + x + 1) dx
= ∫ (x + 1)/√[(x + 1/2)² + 3/4] dx
令x + 1/2 = (√3/2)tanz,dx = (√3/2)sec²z dz
tanz = (2x + 1)/√3,secz = √[1 + (4x² + 4x + 1)/3] = 2√(x² + x + 1)/√3
= ∫ [(√3/2)tanz - 1/2 + 1]/|(√3/2)secz| * (√3/2)sec²z dz
= ∫ [(√3/2)tanz + 1/2] * secz dz
= (1/2)∫ (√3secztanz + secz) dz
= (√3/2)secz + (1/2)ln|secz + tanz| + 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