Question

请问这两题怎么解，帮忙解答一下过程。

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Answer 1

(1)
用拆项公式的待定系数法
令1/[(1 + 2x)(1 + x²)] = A/(1 + 2x) + (Bx + C)/(1 + x²)
则1 = A(1 + x²) + (Bx + C)(1 + 2x)
1 = (A + 2B)x² + (B + 2C)x + (A + C)
A + 2B = 0、B + 2C = 0、A + C = 1
解得A = 4/5、B = - 2/5、C = 1/5
原式 = (4/5)∫ dx/(1 + 2x) - (2/5)∫ x/(1 + x²) dx + (1/5)∫ dx/(1 + x²)
= (4/5)(1/2)∫ d(1 + 2x)/(1 + 2x) - (2/5)(1/2)∫ d(1 + x²)/(1 + x^2) + (1/5)∫ dx/(1 + x²)
= (2/5)ln|1 + 2x| - (1/5)ln(1 + x²) + (1/5)arctan(x) + C

(2)