第1个回答 2023-12-27
∫ x^(1/2)/[ 1+x^(4/3)] dx
u^5 = u^2.(1+u^3 ) -u^2
let
u=x^(1/4)
du = (1/4)x^(-3/4) dx
dx = 4u^3 du
∫ x^(1/2)/[ 1+x^(4/3)] dx
=∫ [u^2/(1+u^3)] [ 4u^3 du]
=4∫ [u^5/(1+u^3)] du
=4∫ [u^2 - u^2/(1+u^3)] du
= 4 [ (1/3)u^3 -(1/3)ln|1+u^3| ] + C
= 4 [ (1/3)x^(3/4) -(1/3)ln|1+x^(3/4)| ] + C