题目

x^（1／2）是分子,分母是x^（1／2）-x^（1／3）,求该式子的不定积分

提问时间：2021-03-27

答案

令u = x^(1/6),x = u⁶,dx = 6u⁵ du
∫ x^(1/3)/[x^(1/2) - x^(1/3)] dx
= ∫ u²/(u³ - u²) * (6u⁵ du)
= 6∫ u⁷/(u³ - u²) du
= 6∫ u⁵/(u - 1) du
而u⁵ = u⁴[(u - 1) + 1]
= u⁴(u - 1) + u⁴
= u⁴(u - 1) + u³[(u - 1) + 1]
= u⁴(u - 1) + u³(u - 1) + u³
= u⁴(u - 1) + u³(u - 1) + u²[(u - 1) + 1]
= u⁴(u - 1) + u³(u - 1) + u²(u - 1) + u²
= u⁴(u - 1) + u³(u - 1) + u²(u - 1) + u[(u - 1) + 1]
= u⁴(u - 1) + u³(u - 1) + u²(u - 1) + u(u - 1) + u
= u⁴(u - 1) + u³(u - 1) + u²(u - 1) + u(u - 1) + (u - 1) + 1
所以原式 = 6∫ [u⁴ + u³ + u² + u + 1 + 1/(u - 1)] du
= 6[(1/5)u⁵ + (1/4)u⁴ + (1/3)u³ + (1/2)u² + u + ln|u - 1|] + C
将u = x^(1/6)代回,得结果
= (6/5)x^(5/6) + (3/2)x^(2/3) + x + 2x^(1/2) + 3x^(1/3) + 6x^(1/6) + 6ln|1 - x^(1/6)| + C