计算：不定积分∫1/[(1+2x）(1+x^2)]dx 定积分∫ (x-2) √(4-x^2)dx （上限2,下限-2）

待定系数法：
1/[(1 + 2x)(1 + x^2)] = a/(1 + 2x) + (bx + c)/(1 + x^2)
1 = a(1 + x^2) + (bx + c)(1 + 2x)
当x = -1/2,1 = 1.25a => a = 4/5
当x = 0,1 = 4/5 + c => c = 1/5
当x = 1,1 = 8/5 + (b + 1/5)(3) => b = -2/5
∫ dx/[(1 + 2x)(1 + x^2)]
= (4/5)∫ dx/(1 + 2x) - (2/5)∫ x/(1 + x^2) dx + (1/5)∫ dx/(1 + x^2)
= (4/5)(1/2)ln|1 + 2x| - (2/5)(1/2)ln(1 + x^2) + (1/5)arctan(x) + C
= (2/5)ln|1 + 2x| - (1/5)ln(1 + x^2) + (1/5)arctan(x) + C
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∫(-2到2) (x - 2)√(4 - x^2) dx
= ∫(-2到2) x√(4 - x^2) dx - 2∫(-2到2) √(4 - x^2) dx
= 0 - 4∫(0到2) √(4 - x^2) dx
用几何意义解比较快速
表示的圆是x^2 + y^2 = 4,半径为2,
范围由-2到2表示半圆面积
范围由0到2表示1/4的圆面积
= π(2)² * 1/4 = π
所以定积分 = -4(π) = -4π
方法二：用第二类换元法,用代换x = 2sinθ即可.