1⼀√（1+x∧3）原函数是啥

先分解因式：
∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx
= ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx
1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C
1 = (A + B)x² + (- A + B + C)x + (A + C)
{ A + B = 0
{ - A + B + C = 0
{ A + C = 1
(A + B) - (- A + B + C) = 0 ==> { 2A - C = 0
{A + C = 1
(2A - C) + (A + C) = 1 ==> 3A = 1 ==> A = 1/3
B = - 1/3，C = 2/3
原式 = (1/3)∫ dx/(x + 1) - (1/3)∫ (x - 2)/(x² - x + 1) dx
= (1/3)∫ d(x + 1)/(x + 1) - (1/3)∫ [(2x - 1)/2 - 3/2]/(x² - x + 1) dx
= (1/3)ln(x + 1) - (1/6)∫ d(x² - x + 1)/(x² - x +1) + (1/2)∫ d(x - 1/2)/[(x - 1/2)² + 3/4]
= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/2)(2/√3)arctan[(x - 1/2) * 2/√3] + C
= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/√3)arctan(2x/√3 - 1/√3) + C
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