∫x*根号((1-x)/(1+x))dx求解答

∫x*√[(1-x)/(1+x)]dx
=∫ [x(1-x)/√ (1-x^2) ] dx
let
x= siny
dy = cosydy
∫ [x(1-x)/√ (1-x^2) ] dx
=∫ siny(1-siny) dy
= ∫ [siny + (cos2y -1)/2] dy
= -cosy + sin2y/4 - y/2 + C
= -√ (1-x^2) + x√ (1-x^2)/2 - (arcsinx)/2 + C