不定积分dx/(1-x^1/3)

Answer 1

答：
设t=³√x，x=t³
原式=∫ 1/(1-³√x) dx
=∫ 1/(1-t) d(t³)
=∫ 3t² /(1-t) dt
=3∫ (1-t-1)² /(1-t) dt
=3 ∫ [ (1-t)²-2(1-t) +1 ] /(1-t ) dt
=3 ∫ (1-t) -2+1/(1-t) dt
=3 ∫ (-1-t) -1/(t-1) dt
=3 [-t-t²/2 -ln|t-1| ]+ C
=-3t -1.5t²-3ln|t-1| +C
=-3³√x-1.5³√x² -3ln|³√x-1|+C

Answer 2

追答