待定系數(shù)法：
1/[(1 + 2x)(1 + x^2)] = a/(1 + 2x) + (bx + c)/(1 + x^2)
1 = a(1 + x^2) + (bx + c)(1 + 2x)
當(dāng)x = -1/2,1 = 1.25a => a = 4/5
當(dāng)x = 0,1 = 4/5 + c => c = 1/5
當(dāng)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
用幾何意義解比較快速
表示的圓是x^2 + y^2 = 4,半徑為2,
范圍由-2到2表示半圓面積
范圍由0到2表示1/4的圓面積
= π(2)² * 1/4 = π
所以定積分 = -4(π) = -4π
方法二：用第二類換元法,用代換x = 2sinθ即可.