令g(x)=f(x)+f(1/x)，x>0g(x)=arctan(1/x)+arctanxg'(x)=[1/(1+1/x^2)]*(-1/x^2)+1/(1+x^2)=0所以g(x)=C，其中C是任意常数因为g(1)=arctan1+arctan1=π/4+π/4=π/2所以C=π/2即g(x)=π/2f(x)+f(1/x)=π/2
