高等数学不定积分第6题，求大神解答
展开全部
6. 设(x^2+1)/[(x-1)(x+1)^2] = A/(x-1) + B/(x+1) + C/(x+1)^2
则 (x^2+1)/[(x-1)(x+1)^2] = (Ax^2+2Ax+A+Bx^2-B+Cx-C)/[(x-1)(x+1)^2]
得 A+B = 1, 2A+C = 0, A-B-C = 1,
联立解得 A = B = 1/2, C = -1,
则 I = (1/2)∫dx/(x-1) + (1/2)∫dx/(x+1) - ∫dx/(x+1)^2
= (1/2)ln|x^2-1| + 1/(x+1) + C
则 (x^2+1)/[(x-1)(x+1)^2] = (Ax^2+2Ax+A+Bx^2-B+Cx-C)/[(x-1)(x+1)^2]
得 A+B = 1, 2A+C = 0, A-B-C = 1,
联立解得 A = B = 1/2, C = -1,
则 I = (1/2)∫dx/(x-1) + (1/2)∫dx/(x+1) - ∫dx/(x+1)^2
= (1/2)ln|x^2-1| + 1/(x+1) + C
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
