就用基本的积分公式来解即可
∫[-1,1](x+√4-x^2)^2dx
=∫[-1,1](x^2+4-x^2 +2x*√4-x^2) dx
=∫[-1,1] 4+2x*√4-x^2dx
=4x - 2/3 *(4-x^2)^(3/2) 代入上下限1和-1
=8
∫[-1,1](x+√4-x^2)^2dx
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