原式=∫[-1,1] (x+√(1-x^2))^2 dx ∫[-1,1]表示-1到1的定积分=∫[-1,1] x^2 dx + ∫[-1,1] (1-x^2) dx +2∫[-1,1] x√(1-x^2) dx=∫[-1,1] dx +∫[-1,1] √(1-x^2) dx^2=2-∫[-1,1] √(1-x^2) d(1-x^2) =2- (2/3) *...