Question

(x^2 1)/(x^4 1)的不定积分

(x^2+1)/(x^4+1)不知道为什么提交后加号没了

Answer 1

= ∫ d(x^2) / 2 / (1+ x^2)^1/2
= (1+ x^2)^1/2 + C

Answer 2

一楼的结果不对，那个结果求导后为：x²/(x²-1)²

本题解法技巧较高
∫ x²/(1+x^4) dx
=(1/2)∫ (x²-1+x²+1)/(1+x^4) dx
=(1/2)∫ (x²-1)/(1+x^4) dx + (1/2)∫ (x²+1)/(1+x^4) dx
分子分同除以x²
=(1/2)∫ (1-1/x²)/(1/x²+x²) dx + (1/2)∫ (1+1/x²)/(1/x²+x²) dx
分子放到微分之后
=(1/2)∫ 1/(1/x²+x²) d(x+1/x) + (1/2)∫ 1/(1/x²+x²) d(x-1/x)
=(1/2)∫ 1/(1/x²+x²+2-2) d(x+1/x) + (1/2)∫ 1/(1/x²+x²-2+2) d(x-1/x)
=(1/2)∫ 1/[(x+1/x)²-2] d(x+1/x) + (1/2)∫ 1/[(x-1/x)²+2] d(x-1/x)
=(√2/8)ln|(x+1/x-√2)/(x+1/x+√2)| + (√2/4)arctan[(x-1/x)/√2] + C
=(√2/8)ln|(x²+1-√2x)/(x²+1+√2x)| + (√2/4)arctan[(x-1/x)/√2] + C

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Answer 3

令x=tany
∫(x^2/(1+x^4))dx
=∫(tany^2/(1+tany^4))*(1/(cosy)^2)dy
=∫(siny)^2/((siny)^4+(cosy)^4) dy
=∫(1/2)(1-cos2y)/(1-4(siny)^2(cosy)^2) dy
=(1/2)∫(1-cos2y)/(1-(sin2y)^2) dy
=(1/2)∫1/(1-(sin2y)^2) dy - (1/2)∫cos2y/(1-(sin2y)^2) dy
=(1/4)∫(1/(cos2y)^2)d(2y) - (1/4)∫1/((1-sin2y)(1+sin2y)) d(sin2y)
=(1/4)tan2y - (1/8)∫(1/(1-sin2y) + 1/(1+sin2y))d(sin2y)
=(1/4)tan2y - (1/8)ln((1+sin2y)/(1-sin2y)) + C
=(1/4)tan2y - (1/4)ln|(siny+cosy)/(siny-cosy)| + C
=(1/2)tany/(1-(tany)^2) - (1/4)ln|(tany+1)/(tany-1)| + C
=(1/2)x/(1-x^2) - (1/4)ln|(x+1)/(x-1)| + C