Question

有理函数的积分法怎么拆项

Answer 1

(2) 积分函数 f(x) = (x^2+1)/[(x-1)(x+1)^2] 用待定系数法，设分拆成以下有理分式 f(x) = A/(x-1) + B/(x+1) + C/(x+1)^2 通分得 f(x) = [A(x+1)^2 + B(x+1)(x-1) + C(x-1)] / [(x-1)(x+1)^2] = [(A+B)x^2 + (2A+C)x + (A-B-C)] / [(x-1)(x+1)^2] 与原式比较，分母同，分子中 x 同次幂的系数必然相同，得 A+B = 1, 2A+C = 0, A-B-C = 1, 联立解得 A = B = 1/2, C = -1, 则 f(x) = (1/2)[1/(x-1) + 1/(x+1)] - 1/(x+1)^2 另题简单，仿做即可。

追问

能写在纸上吗 谢谢