设(x-1)/(x+1)=t³,则dx=6t²dt/(1-t³)²，x+1=2/(1-t³),x-1=2t³/(1-t³)
∴原式=∫[6t²dt/(1-t³)²]/{[2/(1-t³)]^(2/3)*[2t³/(1-t³)]^(4/3)}
=∫(6t²dt)/[2²t^4]
=3/2∫dt/t²
=(-3/2)/t+C
=(-3/2)[(x+1)/(x-1)]^(1/3)+C (C是积分常数)
或者
