∫dx/(根号(2x+1)+根号(2x-1))
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/08/27 01:35:57
∫dx/(根号(2x+1)+根号(2x-1))
∫1/[√(2x+1)+√(2x-1)] dx
= ∫1/[√(2x+1)+√(2x-1)] * [√(2x+1)-√(2x-1)]/[√(2x+1)-√(2x-1)] dx,分母有理化
= (1/2)∫[√(2x+1)-√(2x-1)] dx
= (1/2)∫√(2x+1) dx - (1/2)∫√(2x-1) dx
= (1/4)∫√(2x+1) d(2x+1) - (1/4)∫√(2x-1) d(2x-1)
= (1/4)(2/3)(2x+1)^(3/2) - (1/4)(2/3)(2x-1)^(3/2) + C
= (1/6) * [(2x+1)^(3/2) - (2x-1)^(3/2)] + C
= ∫1/[√(2x+1)+√(2x-1)] * [√(2x+1)-√(2x-1)]/[√(2x+1)-√(2x-1)] dx,分母有理化
= (1/2)∫[√(2x+1)-√(2x-1)] dx
= (1/2)∫√(2x+1) dx - (1/2)∫√(2x-1) dx
= (1/4)∫√(2x+1) d(2x+1) - (1/4)∫√(2x-1) d(2x-1)
= (1/4)(2/3)(2x+1)^(3/2) - (1/4)(2/3)(2x-1)^(3/2) + C
= (1/6) * [(2x+1)^(3/2) - (2x-1)^(3/2)] + C