求不定积分√{√[1-(x-1)²]-x}dx原式=∫√[1-(x-1)²]dx-∫xdx=∫√[1-(x-1)²]d(x-1)-x²/2令x-1=sinu,则d(x-1)=cosudu,故其中∫√[1-(x-1)²]d(x-1)=∫cos²udu=∫[(1+cos2u)/2]du=u/2+(...sorry可能是写的不清是f10太谢谢了