∫xdx/√1+x^(2/3) 求高人指点
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令u = x^(2/3)，x = u^(3/2)，dx = (3/2)√u du
∫ x/√[1 + x^(2/3)] dx
= ∫ u^(3/2)/√(1 + u) · (3/2)√u du
= (3/2)∫ u²/√(1 + u) du
令s² = 1 + u，2sds = du
= (3/2)∫ (s² - 1)²/s · 2sds
= 3∫ (s⁴ - 2s² + 1) ds
= (3/5)s⁵ - 2s³ + 3s + C
= (3/5)(1 + u)^(5/2) - 2(1 + u)^(3/2) + 3√(1 + u) + C
= (1/5)[3x^(4/3) - 4x^(2/3) + 8]√[1 + x^(2/3)] + C
∫ x/√[1 + x^(2/3)] dx
= ∫ u^(3/2)/√(1 + u) · (3/2)√u du
= (3/2)∫ u²/√(1 + u) du
令s² = 1 + u，2sds = du
= (3/2)∫ (s² - 1)²/s · 2sds
= 3∫ (s⁴ - 2s² + 1) ds
= (3/5)s⁵ - 2s³ + 3s + C
= (3/5)(1 + u)^(5/2) - 2(1 + u)^(3/2) + 3√(1 + u) + C
= (1/5)[3x^(4/3) - 4x^(2/3) + 8]√[1 + x^(2/3)] + C
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