Question

求1/(√x+³√x)不定积分

无

Answer 1

简单计算一下，答案如图所示

Answer 2

u^3
=u^2.(u+1) -u^2
=u^2.(u+1) -u(u+1) +u
=u^2.(u+1) -u(u+1) +(u+1) -1
u^3/(u+1) = u^2 -u +1 -1/(u+1)
let
u= x^(1/6)
du =(1/6)x^(-5/6) dx
dx =6u^5 du
∫ dx/[√x+x^(1/3)]
=∫ 6u^5 du/(u^3+u^2)
=6∫ u^3/(u+1) du
=6∫[ u^2 -u +1 -1/(u+1)] du
=6[ (1/3)u^3 -(1/2)u^2 + u - ln|u+1| ] +C
=2u^3 -3u^2 + 6u - 6ln|u+1| +C
=2x^(1/2) -3x^(1/6) + 6x^(1/6) -6ln|x^(1/6) +1| +C