求(x4+1)/(x6+1)的不定积分

先将分母x^6+1分解为
3个二次式因子(xx+1)(xx+1-√3x)(xx+1+√3x),
用待定系数法求出
(x^4+1)/(x^6+1)
=A/(xx+1)+B/(xx+1-√3x)+B/(xx+1+√3x)
A=2/3,B=1/6
∫A/(xx+1)dx=2/3arctg(x)+C
4(xx+1±√3x)=(2x±√3)^2+1,
作变量代换t=2x±√3,
则 ∫B/(xx+1±√3x)dx
=1/3∫1/(tt+1)dt
=1/3arctg(t)+C
=1/3arctg(2x±√3)+C
所以
∫(x^4+1)/(x^6+1)dx
=[2arctgx+arctg(2x-√3)+arctg(2x+√3)]/3+C
arctg(2x-√3)+arctg(2x+√3)=arctg[x/(1-xx)]
所以
∫(x^4+1)/(x^6+1)dx
=2/3·arctgx+1/3·arctg[x/(1-xx)]+C