Question

积分题如图

Answer 1

追问

这为什么会相等

第二行的那步怎么推的，谢谢

提出1/2的那步怎么做

追答

做错了

这样才对，抱歉了

追问

谢谢谢谢

Answer 2

x^4 + x^2 + 1 = (x^2+1)^2 - x^2 = (x^2+x+1)(x^2-x+1)
设 1/(x^4+x^2+1) = (ax+b)/(x^2+x+1) + (cx+d)/(x^2-x+1)
= [(ax+b)(x^2-x+1)+(cx+d)(x^2+x+1)]/[(x^2+x+1)(x^2-x+1)]
则 a+c = 0， b-a+d+c = 0, a-b+c+d = 0, b+d = 1
联立解得 a = b = d = 1/2， c = -1/2，
1/(x^4+x^2+1) = (1/2)[(x+1)/(x^2+x+1) - (x-1)/(x^2-x+1)]
则 I = ∫dx/(x^4+x^2+1)
= (1/2)∫(x+1)dx/(x^2+x+1) - (1/2)∫(x-1)dx/(x^2-x+1)
= (1/4)∫(2x+1+1)dx/(x^2+x+1) - (1/4)∫(2x-1-1)dx/(x^2-x+1)
= (1/4)ln(x^2+x+1) + (1/4)∫d(x+1/2)/[(x+1/2)^2+3/4]
- (1/4)ln(x^2-x+1) + (1/4)∫d(x-1/2)/[(x-1/2)^2+3/4]
= (1/4)ln[(x^2+x+1)/(x^2-x+1)] + (1/4)arctan[(2x+1)/√3]
+ (1/4)arctan[(2x-1)/√3] + C