这个好像是一个公式
let x^(3/8) = tanu (3/8)x^(-5/8) dx = (secu)^2 du x^(1/2) dx = (8/3) (tanu)^3. (secu)^2 du / ∫ x^(1/2)/[ x^(3/4) +1 ] dx =∫ (8/3) (tanu)^3. (secu)^2 du/ (secu)^2 =(8/3) ∫ (tanu)^3 du =(8/3) ∫ (tanu)[( secu)^2 -1 ] du =(8/3) ∫ tanudtanu - (8/3) ∫ tanu du =(4/3) (tanu)^2+ (8/3)ln|cosx| + C =(4/3) x^(3/4)+ (8/3)ln| 1/√[x^(3/4) +1]| + C =(4/3) x^(3/4)- (4/3)ln| x^(3/4) +1| + C