一样是分解成部分分式,通常用待定系数法：比如设3/(x^3+1)=a/(x+1)+(bx+c)/(x^2-x+1)则3=a(x^2-x+1)+(bx+c)(x+1)(a+b)x^2+(b+c-a)x+a+c=3对比系数：a+b=0, b+c-a=0, a+c=3得：a=1, b=-1, c=2故有3/(x^3+1)=1/(x+1)+(...