∫ [x+√(1-x²)]²dx=∫ x²+1-x²+2x√(1-x²)dx=∫ 1+√(1-x²)d(x²)=x+(1-x²)^(3/2)*2/3+C
则∫(1 -1) [x+√(1-x²)]²dx=1-(-1)=2