设：(e^x-1)^(1/3)=y e^x-1=y^3 e^x=1+y^3 e^xdx=3y^2dy ∫(2,0)e^xdx/(e^x-1)^(1/3) =3∫[(e^2-1)^(1/3),0] y^2dy/y=3∫[(e^2-1)^(1/3),0] ydy=1.5 y^2 | [(e^2-1)^(1/3),0]= 3[(e^2-1)^(2/3)]/2