Question

1/（1+2x）（1+x^2）的不定积分

Answer 1

S(1/(1+2x)(1+x^2))dx
=S(4/5*1/(1+2x)-1/5*(2x-1)/(x^2+1))dx
=2/5*ln(1+2x)-1/5*ln(x^2+1)-1/5*arctanx+c

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Answer 2

令1/[(1 + 2x)(1 + x^2)] = A/(1 + 2x) + (Bx + C)/(1 + x^2)
则1 = A(1 + x^2) + (Bx + C)(1 + 2x)
1 = (A + 2B)x^2 + (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 + x^2) dx + (1/5)∫ dx/(1 + x^2)
= (4/5)(1/2)∫ d(1 + 2x)/(1 + 2x) - (2/5)(1/2)∫ d(1 + x^2)/(1 + x^2) + (1/5)∫ dx/(1 + x^2)
= (2/5)ln|1 + 2x| - (1/5)ln(1 + x^2) + (1/5)arctan(x) + C

Answer 3