Question

求∫ x/(1+√x) dx 的积分。

Answer 1

令u = √x，x = u²，dx = 2u du
∫ x/(1 + √x) dx
= ∫ u²/(1 + u) * (2u du)
= 2∫ u²[(1 + u) - 1]/(1 + u) du
= 2∫ u² du - 2∫ u²/(1 + u) du
= (2/3)u³ - 2∫ u[(u + 1) - 1]/(1 + u) du
= (2/3)u³ - 2∫ u du + 2∫ u/(1 + u) du
= (2/3)u³ - u² + 2∫ [(u + 1) - 1]/(1 + u) du
= (2/3)u³ - u² + 2∫ du - 2∫ du/(1 + u)
= (2/3)u³ - u² + 2u - 2ln(1 + u) + C
= (2/3)x^(3/2) - x + 2√x - 2ln(1 + √x) + C