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
原式=∫(0→2)d(e^x)/(e^x-1)^(1/3)=3/2(e^x-1)^(2/3)|(0→2)=3/2(e^2-1)^(2/3)
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024