换元法求积分。如图所示:
供参考，请笑纳。
Let u = x^(1/4)x = u^4dx = 4u^3 du原积分 = ∫4u^3/[(1+u)^3 u^2 du= 4∫[(u+1)-1]/[(1+u)^3 du= 4∫1/(1+u)^2 - 1/(1+u)^3 du= 4[-1/(1+ x^(1/4)) + 0.5/(1+ x^(1/4))^2] + c
