1/(x^3+1) = 1/(1+x+1) / (x^2-x+1) = A/(x+1) + (Bx+C)/(x^2-x+1)
待定系數(shù)法 => A=1/3,B=-1/3,C=2/3
∫(1/(x^3+1))dx
= (1/3) ln(x+1) – (1/6) ∫ (2x-1) / (x^2-x+1) dx + (1/2) ∫ dx / (x^2-x+1)
= (1/3) ln(x+1) – (1/6) ln (x^2-x+1) + (1/2)(2/√3)arctan(2x-1)/√3 + C
= (1/3) ln(x+1) – (1/6) ln (x^2-x+1) + (1/√3)arctan(2x-1)/√3 + C1/(x^3+1) = 1/(1+x+1) / (x^2-x+1) = A/(x+1) + (Bx+C)/(x^2-x+1)這步我不懂啊，為什么不是1/(x^3+1) = 1/(1+x+1) / (x^2-x+1) = A/(x+1) + (B)/(x^2-x+1)突然冒出一個(Bx+C）不懂為什么，能解釋下嗎更正： x^3+1 =(1+x) *(x^2-x+1) 設 1/(x^3+1) = 1/(1+x) / (x^2-x+1) =A/(x+1) + (Bx+C)/(x^2-x+1)有理函數(shù)需要分解成部分分式之和。(Bx+C)/(x^2-x+1) 當分母是二次式，分子需設為一次多項式。