=arctg •sinπ−cosπ•ln(x2−2x cosπ+1)+C
解：∵ x2n+1=x2n−ei(2N+1)π=x2n−(eiπ)2n，则x2n+1必有因式x−eiπ，N为任意整数．
∴ x2n+1的因式集为{x−eiπ|k=1,2,…,2n.}，取a=eiπ，
x2n+1=(x−eiπ)=(x−a2k−1)=(x−a1)(x−a3)…(x−a2k−1)…(x−a4n−1)．
则 x2n+1可表述为：x2n+1=(x−a2k1−1)(x−a2k2−1)…(x−a2kl−1)…(x−a2k2n−1)，(2k−1)=(2kl−1)．
∴ x2n+1的x2n−l项系数为：(−a2k1−1)(−a2k2−1)…(−a2kl−1)
=(−1)l•a2k1−1•a2k2−1•…•a2k2n−1=(−1)l• a2kj−1=0．
x2n+1的常数项为：(−eiπ)=(−a2kl−1)=a2k−1=a2k−1•a−(2k−1)=1=1．
∴ = = •(x−a1)(x−a3)…(x−a2k−1)…(x−a4n−1)
=(x−a2k1−1)(x−a2k2−1)…(0−a2kl−1)…(x−a2k2n−1)=
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