既然满足三次方根x,又要满足二次方根x,当然是它们的最小公倍数6
令x = u^6,dx = 6u^5 du
∫ dx / (x^1/3 + x^1/2) dx
= ∫ 6u^5 / [(u^6)^1/3 + (u^6)^1/2] du
= 6∫ u^5 / (u² + u³) du
= 6∫ u³ / (1 + u) du
= 6∫ [u² - u - 1/(u+1) + 1] du
= 6[u³/3 - u²/2 - ln|u+1| + u] + C
= 2u³ - 3u² - 6ln|1+u| + 6u + C
= 2√x - 3x^(1/3) + 6x^(1/6) - 6ln|1+x^(1/6)| + C
PS：x^(1/n)是n次方根x,所以x^(1/6)是6次方根x