求积分x^1/2/(1-x^1/3)dx

网友回答

令a=x^1/6
则x=a^6
dx=6a^5da
原式=∫a³/(1-a²)*6a^5da
=6∫a^8/(1-a²)da
=6∫(a^8-1+1)/(1-a²)da
=6∫[(a^4+1)(a²+1)(a²-1)/(1-a²)+1/(1-a²)]da
=6∫[-a^6-a^4-a²-1+1/(1-a²)]da
=-6x^7/7-6a^5/5-2a³-6a+ln|(1-a)/(1-a)|+C
再把a代回去即可