∫dx/(1+x)(1+x^2)=?
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解决时间 2021-02-14 12:07
- 提问者网友：浮克旳回音
- 2021-02-13 11:29
∫dx/(1+x)(1+x^2)=?
最佳答案
- 五星知识达人网友：罪歌
- 2021-02-13 12:20
可用待定系数法。
令1/[(1 + x)(1 + x^2)] = A/(1 + x) + (Bx + C)/(1 + x^2)
==> 1 = A(1 + x^2) + (Bx + C)(1 + x)
==> 1 = (A + B)x^2 + (B + C)x + (A + C)
∴A + B = 0 ==> B = - A
∴B + C = 0 ==> C = - B
∴A + C = 1 ==> C = 1 - A
有1 - A = - (- A) ==> A = 1/2、B = - 1/2、C = 1/2
于是∫ 1/[(1 + x)(1 + x^2)] dx
= (1/2)∫ 1/(1 + x) dx - (1/2)∫ x/(1 + x^2) dx + (1/2)∫ 1/(1 + x^2) dx
= (1/2)ln|1 + x| - (1/4)ln(1 + x^2) + (1/2)arctan(x) + C
= (1/4)ln[(1 + x)^2/(1 + x^2)] + (1/2)arctan(x) + C
令1/[(1 + x)(1 + x^2)] = A/(1 + x) + (Bx + C)/(1 + x^2)
==> 1 = A(1 + x^2) + (Bx + C)(1 + x)
==> 1 = (A + B)x^2 + (B + C)x + (A + C)
∴A + B = 0 ==> B = - A
∴B + C = 0 ==> C = - B
∴A + C = 1 ==> C = 1 - A
有1 - A = - (- A) ==> A = 1/2、B = - 1/2、C = 1/2
于是∫ 1/[(1 + x)(1 + x^2)] dx
= (1/2)∫ 1/(1 + x) dx - (1/2)∫ x/(1 + x^2) dx + (1/2)∫ 1/(1 + x^2) dx
= (1/2)ln|1 + x| - (1/4)ln(1 + x^2) + (1/2)arctan(x) + C
= (1/4)ln[(1 + x)^2/(1 + x^2)] + (1/2)arctan(x) + C
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