先計算
∫ x³/√(1-x²) dx
=(1/2)∫ x²/√(1-x²) d(x²)
令√(1-x²)=u,則x²=1-u²,d(x²)=-2udu
=(1/2)∫ [(1-u²)/u](-2u) du
=∫ (u²-1) du
=(1/3)u³ - u + C
=(1/3)(1-x²)^(3/2) - √(1-x²) + C

因此：x³/√(1-x²) dx = d[(1/3)(1-x²)^(3/2) - √(1-x²)]
下面計算本題

∫ x³arccosx/√(1-x²) dx
=∫ arccosx d[(1/3)(1-x²)^(3/2) - √(1-x²)]
分部積分
=(1/3)(1-x²)^(3/2)arccosx - √(1-x²)arccosx - ∫ [(1/3)(1-x²)^(3/2) - √(1-x²)] darccosx
=(1/3)(1-x²)^(3/2)arccosx - √(1-x²)arccosx + ∫ [(1/3)(1-x²) - 1] dx
=(1/3)(1-x²)^(3/2)arccosx - √(1-x²)arccosx - (2/3)x - (1/9)x³ + C

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