积分1/x^4(x^2+1)dx请高手帮忙把过程写出来 谢谢

网友回答

可惜,楼上解繁了,也解错了.
利用待定系数法：
∫dx/(x²+1)x⁴
=∫[A/x + B/x² + C/x³ + D/x⁴+ (Ex + F)/(x² + 1)]dx
=∫[-1/x² + 1/x⁴+ 1/(x² + 1)]dx
= 1/x - 1/3x³ + arctanx + C
======以下答案可供参考======
供参考答案1:
令: x=tanu , dx=sec^2 udu ,cscu=x/√(1+x^2)
∫1/[x^4(x^2+1)]dx
= ∫1/(tan^4 u*sec^2 u)*sec^2 u du
= ∫1/tan^4 u *du
= ∫cot^4 u *du
= ∫cot^2 u *(cot^2 du)
= -∫(csc^2 u -1) dcscu
= cscu -1/3*csc^3 u +C
=x/√(1+x^2) -[x/√(1+x^2)]^3/3 + C