Question

求大神指教，求这个不定积分

Answer 1

如图

Answer 2

令a=x^(1/6)
则x=a^6
³√x=a²，√x=a³
dx=6a^da
所以原式=∫6a^da/a³(1+a²)
=∫6a²da/(1+a²)
=6∫(1+a²-1)da/(1+a²)
=6∫[1-1/(1+a²)]da
=6a-6arctana+C
=6x^(1/6)-6arctan[x^(1/6)]+C