Question

（x^4+x^2+1）分之一的不定积分怎么求？

急急急……

Answer 1

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！

Answer 2

∫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