∫dx/(根号(2x+1)+根号(2x-1))
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∫dx/(根号(2x+1)+根号(2x-1))
![∫dx/(根号(2x+1)+根号(2x-1))](/uploads/image/z/19401998-14-8.jpg?t=%E2%88%ABdx%2F%28%E6%A0%B9%E5%8F%B7%282x%2B1%29%2B%E6%A0%B9%E5%8F%B7%282x-1%29%29)
∫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