第1个回答 2011-11-30
提供一个思路吧
x^4+x^2+1 = x^4 + 2x^2 + 1 - x^2 = (x^2+1)^2-x^2 = (x^2-x+1)(x^2+x+1)
这样就可以利用部分分式法把原来的积分转化为两个有理分式的积分
1/(x^4+x^2+1) = (Ax+B)/(x^2-x+1) + (Cx+D)/(x^2+x+1) A,B,C,D可以通过待定系数法确定
有理分式的积分方法相信书本上已经讲的很全面了，后面的你就自己算吧
仅供参考！希望可以帮到你！本回答被提问者采纳
x^4+x^2+1 = x^4 + 2x^2 + 1 - x^2 = (x^2+1)^2-x^2 = (x^2-x+1)(x^2+x+1)
这样就可以利用部分分式法把原来的积分转化为两个有理分式的积分
1/(x^4+x^2+1) = (Ax+B)/(x^2-x+1) + (Cx+D)/(x^2+x+1) A,B,C,D可以通过待定系数法确定
有理分式的积分方法相信书本上已经讲的很全面了，后面的你就自己算吧
仅供参考！希望可以帮到你！本回答被提问者采纳
第2个回答 2011-11-30
∫1/(x⁴+x²+1) dx
= (1/2)∫(x+1)/(x²+x+1) dx - (1/2)∫(x-1)/(x²-x+1) dx
= (1/2)[(1/2)∫(2x+1)/(x²+x+1) dx + (1/2)∫1/(x²+x+1) dx]
- (1/2)[(1/2)∫(2x-1)/(x²-x+1) dx - (1/2)∫1/(x²-x+1) dx]
= (1/4)∫d(x²+x+1)/(x²+x+1) - (1/4)∫d(x²-x+1)/(x²-x+1)
+ (1/4)∫d(x+1/2)/[(x+1/2)²+3/4] + (1/4)∫d(x-1/2)/[(x-1/2)²+3/4]
= (1/4)ln|(x²+x+1)/(x²-x+1)| + (2/√3)arctan[(2x+1)/√3] + (2/√3)arctan[(2x-1)√3] + C
= (1/2)∫(x+1)/(x²+x+1) dx - (1/2)∫(x-1)/(x²-x+1) dx
= (1/2)[(1/2)∫(2x+1)/(x²+x+1) dx + (1/2)∫1/(x²+x+1) dx]
- (1/2)[(1/2)∫(2x-1)/(x²-x+1) dx - (1/2)∫1/(x²-x+1) dx]
= (1/4)∫d(x²+x+1)/(x²+x+1) - (1/4)∫d(x²-x+1)/(x²-x+1)
+ (1/4)∫d(x+1/2)/[(x+1/2)²+3/4] + (1/4)∫d(x-1/2)/[(x-1/2)²+3/4]
= (1/4)ln|(x²+x+1)/(x²-x+1)| + (2/√3)arctan[(2x+1)/√3] + (2/√3)arctan[(2x-1)√3] + C
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