求不定积分(1+x^2)/(1+x^6)dx

求不定积分(1+x^2)/(1+x^6)dx
求不定积分(1+x^2)/(1+x^6)dx

求不定积分(1+x^2)/(1+x^6)dx
(1+x^2)/(1+x^6)=1/(x^4-x^2+1)
=1/[x^2+3^(1/2)x+1]*[x^2-3^(1/2)x+1]
设原式=(ax+b)/[x^2+3^(1/2)x+1]
+(cx+d)/[x^2-3^(1/2)x+1]
a+c=0 b+d+3^(1/2)(c-a)=0 a+c+3^(1/2)(d-b)=0
b+d=1
则a=3^(1/2)/6 c=-3^(1/2)/6 b=d=1/2
原式=[3^(1/2)/12]*{[2x+3^(1/2)]/[x^2+3^(1/2)x+1]
-[2x-3^(1/2)]/[x^2-3^(1/2)x+1]}+
1/{[2x+3^(1/2)]^2+1}+1/{[2x-3^(1/2)]^2+1}
∴∫(1+x^2)dx/(1+x^6)
=[3^(1/2)/12]In{[x^2+3^(1/2)x+1]/[x^2-3^(1/2)x+1]}
+(1/2){arctan[2x+3^(1/2)]+arctan[2x-3^(1/2)]}+C
=[3^(1/2)/12]In{[x^2+3^(1/2)x+1]/[x^2-3^(1/2)x+1]}
+(1/2)arctan[x/(1-x^2)]+C

所有积分号省略:
令x^2=t
原式=(1+t)/(1+t^3)dt
=1/(1-t+t^2)dt
=1/[3/4+(1/4-t+t^2)]dt
=1/[3/4+(1/2-t)^2]dt
=-1/[3/4+(1/2-t)^2]d(1/2-t)
这一步用公式就行了，再把x回代！