Question

根号下x/1-x的原函数

Answer 1

x-x^2 = 1/4- (x-1/2)^2
let
x-1/2 = (1/2)sinu
dx= (1/2)cosu du
∫√[x/(1-x)] dx
=∫ x/√(x-x^2) dx
=∫ [ (1/2)+(1/2)sinu ] du
= (1/2)u -(1/2)cosu + C
= (1/2)arcsin(2x-1) -√(x-x^2) + C