求不定积分1/[x^2*根号下(x^2+4)]

Question

求不定积分
1/[x^2*根号下(x^2+4)]

Answer 1

都做得不对.
方法一：倒代换
x = 1/z,dx = -1/z² dz
∫ dx/[x²√(x²+4)]
= ∫ (-1/z² dz)/[(1/z²)√(1/z² + 4)]
= -∫ z/√(1+4z²) dz
= (-1/2)∫ 1/√(1+4z²) d(z²)
= (-1/8)∫ 1/√(1+4z²) d(1+4z²)
= (-1/8) * 2√(1+4z²) + C
= (-1/4)√(1+4/x²) + C
= [-√(x²+4)]/(4x) + C
方法二：第二类换元法
∫ dx/[x²√(x²+4)]
(x = 2tanz,dx = 2sec²zdz)
= ∫ (2sec²zdz)/(4tan²z*2secz)
= (1/4)∫ secz/tan²z dz
= (1/4)∫ (cosz/sin²z) dz
= (1/4)∫ csczcotz dz
= (-1/4)cscz + C
= [-√(x²+4)]/(4x) + C
因为tanz = x/2,所以sinz = x/√(x²+2²) = x/√(x²+4),cscz = √(x²+4)/x