令x^(1/3)=t
则原式=∫(0→2)1/(1+t)*3t^2dt
=3∫(0→2)(t^2+t-t-1+1)/(t+1)dt
=3∫(0→2)tdt-3∫(0→2)dt+3∫(0→2)dt/(t+1)
=3/2t^2|(0→2)-3t|(0→2)+3ln|t+1||(0→2)
=6-6++3ln3
=3ln3
用换元法求定积分∫1/(1+x的三次根）dx上限8下限0
令x^(1/3)=t
则原式=∫(0→2)1/(1+t)*3t^2dt
=3∫(0→2)(t^2+t-t-1+1)/(t+1)dt
=3∫(0→2)tdt-3∫(0→2)dt+3∫(0→2)dt/(t+1)
=3/2t^2|(0→2)-3t|(0→2)+3ln|t+1||(0→2)
=6-6++3ln3
=3ln3