Question

求不定积分∫1/[x^(1/2)-x^(2/3)]dx求高手解题要步骤谢谢

Answer 1

原式=∫2x^(1/2)/[x^(1/2)-x^(2/3)]dx^(1/2)
令x^(1/2)=u
原式=∫2u/[u-u^/3]du
=∫2/[1-u^/2]du
=2arctanu+c
将u=x^(1/2)=代入上式
原式=2arctan[x^(1/2)]+c

Answer 2

令 x^(1/6) = t => x = t^6, x^(2/3) = t^4, dx = 6t^5 dt
原积分 = ∫ 1 / [t^3 - t^4] * 6t^5 dt = ∫ 6t^2 / (1 - t) dt
= -6 ∫ [1 + t + 1 / (t - 1)] dt
= -6t - 3t^2 - 6ln|t-1| + C
= -6x^(1/6) - 3x^(1/3) - 6ln|x^(1/6) - 1| + C
C 为常数.

Answer 3

令x^(1/6)=u，dx=6u^5du，代入得：
∫1/[x^(1/2)-x^(2/3)]dx
=∫[1/(u^3-u^4)]6u^5du
=∫[1/(1-u)]6u^2du
=6∫[-u-1+1/(1-u)]du
=-6(u^2/2+u+ln|1-u|+C
=-6(x/2+x^(1/6)+ln|1-x^(1/6)|)+C