方法如下,请作参考:
分享解法如下。令x=t³。∴原式=∫3t²dt/(1+t)=3∫[t-1+1/(1+t)]dt=(3/2)t²-3t+3ln丨t+1丨+C。再将t=x^(1/3)回代即可。
∫[dx/(1+x^(1/3)]【令x^(1/3)=u,则x=u³;dx=3u²du】=3∫[u²du/(1+u)]=3∫[(u-1)+1/(u+1)]du=3[(1/2)u²-u+ln∣u+1∣+C=3[(1/2)x^(2/3)-x^(1/3)+ln∣1+x^(1/3)∣]+C;