求不定积分详解，

2025-05-23 05:00:32

推荐回答（2个）

回答1：

第二类换元积分法：
令x=siny => dx=cosy dy
cosy=√(1-x²)，tany=x/√(1-x²)
∫x³/(1-x²)³ dx
= ∫(sin³y)/(1-sin²y)³ * cosy dy
= ∫(sin³y)/(cos^5y) dy
= ∫tan³ysec²y dy
= ∫tan³y d(tany)
= (1/4)tan⁴y + C
= (1/4)[x/√(1-x²)]⁴ + C
= x⁴/[4(1-x²)²] + C

回答2：

∫x^3/(1-x^2)^3dx
=1/2∫x^2/(1-x^2)^3dx^2
=1/2∫(x^2-1+1)/(1-x^2)^3dx^2
=1/2∫[-1+1/(1-x^2)^3dx^2
=-1/2x^2-1/2∫[1/(1-x^2)^3d(1-x^2)
=-1/2x^2+1/4*(1-x^2)^(-2)+C