由定义域可知0
x(1-x)= x-x^2 = 1/4-(x-1/2)^2letx-1/2 = (1/2)sinudx=(1/2)cosu du∫dx/√[x(1-x)]=∫(1/2)cosu du/[ (1/2)cosu ]=∫ du=u + C=arcsin(2x-1) + C
题目错误
