這題不能不復(fù)雜吧,有難度.那我隔行寫容易看：
∫ (x³+1) / (x³-1) dx
= ∫ (x³+1) / (x-1)(x²+x+1) dx
= ∫ [-2(x+2) / 3(x²+x+1) + 2 / 3(x-1) + 1] dx
= (-2/3)∫ (x+2)/(x²+x+1) dx + (2/3)∫ dx/(x-1) + ∫ dx
= (-2/3)∫ [(2x+1) / 2(x²+x+1) + 3 / 2(x²+x+1)] dx + (2/3)∫ dx/(x-1) + ∫ dx
= (-1/3)∫ (2x+1)dx/(x²+x+1) - ∫ dx/(x²+x+1) + (2/3)∫ dx/(x-1) + ∫ dx
= (-1/3)∫ (2x+1)dx/(x²+x+1) - ∫ dx/[(x+1/2)²+3/4] + (2/3)∫ dx/(x-1) + ∫ dx
= (-1/3)∫ d(x²+x+1)/(x²+x+1) - ∫ d(x+1/2)/[(x+1/2)²+3/4] + (2/3)∫ d(x-1)/(x-1) + ∫ dx
= (-1/3)ln|x²+x+1| - √(4/3)*arctan[√(4/3) * (x+1/2)] + (2/3)ln|x-1| + x + C
= (-1/3)ln|x²+x+1| + (2/3)ln|x-1| + x - (2/√3)arctan[(2x+1)/√3] + C
已經(jīng)是最簡易的答案了.