简单分析一下，答案如图所示
f'(sin^2x)=1-2sin^2x+sin^2x/(1-sin^2x)f'(x)=1-2x+x/(1-x)f(x)=∫[1-2x+x/(1-x)]dx=x-x^2+∫x/(1-x)dx=x-x^2-∫-x/(1-x)dx=x-x^2-∫[1+1/(x-1)]dx=x-x^2-x-ln(1-x)+C=-x^2-ln(1-x)+C
