有理函数的积分求详解？
展开全部
因为1/(x^4+1)=1/(x^4+2x^2+1-2x^2)
=1/[(x^2+1)^2-2x^2]
=1/(x^2+√2x+1)(x^2-√2x+1)
=[(√2/4)*x+1/2]/(x^2+√2x+1)-[(√2/4)*x-1/2]/(x^2-√2x+1)
=(√2/8)*[(2x+2√2)/(x^2+√2x+1)-(2x-2√2)/(x^2-√2x+1)]
=(√2/8)*{(2x+√2)/(x^2+√2x+1)+√2/[(x+√2/2)^2+1/2]-(2x-√2)/(x^2-√2x+1)+√2/[(x-√2/2)^2+1/2]}
所以原式=(√2/8)*∫{(2x+√2)/(x^2+√2x+1)+√2/[(x+√2/2)^2+1/2]-(2x-√2)/(x^2-√2x+1)+√2/[(x-√2/2)^2+1/2]}dx
=(√2/8)*{∫d(x^2+√2x+1)/(x^2+√2x+1)+√2*∫d(x+√2/2)/[(x+√2/2)^2+1/2]-∫d(x^2-√2x+1)/(x^2-√2x+1)+√2*∫d(x-√2/2)/[(x-√2/2)^2+1/2]}
=(√2/8)*[ln(x^2+√2x+1)+2*arctan(√2x+1)-ln(x^2-√2x+1)+2*arctan(√2x-1)]+C，其中C是任意常数
=1/[(x^2+1)^2-2x^2]
=1/(x^2+√2x+1)(x^2-√2x+1)
=[(√2/4)*x+1/2]/(x^2+√2x+1)-[(√2/4)*x-1/2]/(x^2-√2x+1)
=(√2/8)*[(2x+2√2)/(x^2+√2x+1)-(2x-2√2)/(x^2-√2x+1)]
=(√2/8)*{(2x+√2)/(x^2+√2x+1)+√2/[(x+√2/2)^2+1/2]-(2x-√2)/(x^2-√2x+1)+√2/[(x-√2/2)^2+1/2]}
所以原式=(√2/8)*∫{(2x+√2)/(x^2+√2x+1)+√2/[(x+√2/2)^2+1/2]-(2x-√2)/(x^2-√2x+1)+√2/[(x-√2/2)^2+1/2]}dx
=(√2/8)*{∫d(x^2+√2x+1)/(x^2+√2x+1)+√2*∫d(x+√2/2)/[(x+√2/2)^2+1/2]-∫d(x^2-√2x+1)/(x^2-√2x+1)+√2*∫d(x-√2/2)/[(x-√2/2)^2+1/2]}
=(√2/8)*[ln(x^2+√2x+1)+2*arctan(√2x+1)-ln(x^2-√2x+1)+2*arctan(√2x-1)]+C，其中C是任意常数
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
