Question

不定积分求原函数

Answer 1

• 令x=asint→dx=acost·dt

  则：∫√(a²-x²)/x⁴dx

  =∫[(acost)/(asint)⁴·acostdt

  =(1/a²)·∫cos²t·dt/sin⁴t

  =(1/a²)·∫cos²t·csc⁴tdt

  =(1/a²)·∫cot²t·csc²tdt

  =-(1/a²)·∫cot²t·d(cott)

  =-(⅓/a²)·(cot³t)+C

  sint=x/a→cot(t)=√(a²-x²)/x
  

  原积分=-(⅓/a²)·[√(a²-x²)/x]³+C

追问

用倒代换的方法，就是另x=1/t的方法

追答

• 令x=1/t→dx=-dt/t²

  则：∫√(a²-1/t²)/(1/t)⁴·(-dt/t²)

  =-∫t²√(a²-1/t²)dt

  =-∫t√(a²t²-1)dt

  =-(1/2a²)·∫√(a²t²-1)d(a²t²-1)

  =-(1/2a²)·⅔(a²t²-1)^(3/2)+C

  =-(⅓/a²)(a²/x²-1)^(3/2)+C

  =-(⅓/a²)·[√(a²-x²)/x]³+C

追问

谢谢