在未分解分母之前先别用代换法.
令(x + 2)/[(2x + 1)(x² + x + 1)]
= A/(2x + 1) + (Bx + C)/(x² + x + 1)
= [A(x² + x + 1) + (Bx + C)(2x + 1)]/[(2x + 1)(x² + x + 1)]
则x + 2 = (A + 2B)x² + (A + B + 2C)x + (A + C)
A + 2B = 0、A + B + 2C = 1、A + C = 2
解方程得A = 2,B = - 1,C = 0
原题 = 2∫ 1/(2x + 1) dx - ∫ x/(x² + x + 1) dx
= ∫ d(2x + 1)/(2x + 1) - (1/2)∫ [(2x + 1) - 1]/(x² + x + 1) dx
= ln|2x + 1| - (1/2)∫ d(x² + x + 1)/(x² + x + 1) + (1/2)∫ dx/(x² + x + 1)
= ln|2x + 1| - (1/2)ln(x² + x + 1) + (1/2)∫ d(x + 1/2)/[(x + 1/2)² + 3/4]
= ln|2x + 1| - (1/2)ln(x² + x + 1) + (1/2)(2/√3)arctan[(x + 1/2) * 2/√3] + C
= ln|2x + 1| - (1/2)ln(x² + x + 1) + (1/√3)arctan[(2x + 1)/√3] + C
坚持用换元法也可以,不过有些麻烦.
观测分母中的x² + x + 1 = (x + 1/2)² + 3/4
所以你可设x + 1/2 = (√3/2)tanθ,dx = (√3/2)sec²θ dθ
原题 = ∫ [(√3/2)tanθ - 3/2]/[√3tanθ * (3/4)sec²θ] * (√3/2 * sec²θ dθ)
化简做下去可以了