设[(x-1)/(x+1)]^(1/3)=t,则x=(1+t³)/(1-t³),dx=6t²dt/(1-t³)²
原式=∫dx/{(x+1)(x-1)[(x-1)/(x+1)]^(1/3)}
=∫[6t²dt/(1-t³)²]/{[2/(1-t³)][2t³/(1-t³)]t}
=∫6t²dt/(4t^4)
=3/2∫dt/t²
=-3/2/t+C
=-3/2[(x+1)/(x-1)]^(1/3)+C (C是积分常数).
设[(x-1)/(x+1)]^(1/3)=t,则x=(1+t³)/(1-t³),dx=6t²dt/(1-t³)²
原式=∫dx/{(x+1)(x-1)[(x-1)/(x+1)]^(1/3)}
=∫[6t²dt/(1-t³)²]/{[2/(1-t³)][2t³/(1-t³)]t}
=∫6t²dt/(4t^4)
=3/2∫dt/t²
=-3/2/t+C
=-3/2[(x+1)/(x-1)]^(1/3)+C (C是积分常数).