Question

1/(1+x^4)的原函数怎么求?

Answer 1

分解因式：x^4+1=x^4+1+2x^2-2x^2=(x^2+1)^2-2x^2=(x^2+ √2x+1)(x^2- √配清2x+1)
待定系数法部分分式分1/(x^4+1)=(ax+b)/(x^2+ √2x+1)+(cx+d)/培搏前(x^2- √2x+1)
去分母：1=(ax+b)(x^2- √2x+1)+(cx+d)(x^2+ √2x+1)
1=x^3(a+c)+x^2(b+d- √2a+ √2c)+x(a- √2b+c+ √2d)+b+d
对比系数：b+d=1,a+c=0,b+d- √2a+ √2c=0,a- √2b+c+ √2d=0
解得：a= √2/ 4,c=- √2/4,b=d=1/2
1/(x^4+1)= √2/4*[ (x+ √银并2)/(x^2+ √2x+1)+(-x+ √2)/(x^2- √2x+1)]
这样就可以用基本积分公式来得到结果了.