嘿嘿,其实这题很简单.
令y = 1/x、x = 1/y、dx = - 1/y² dy
∫ [arctan(1/x)]/(1 + x²) dx
= ∫ arctany/(1 + 1/y²) * (- 1/y² dy)
= ∫ arctany * y²/(1 + y²) * (- 1/y²) dy
= - ∫ arctany/(1 + y²) dy
= - ∫ arctany d(arctany)
= (- 1/2)(arctany)² + C
= (- 1/2)[arctan(1/x)]² + C