证明 arctanx+arctan1/x=π/2 (x>0) 用中值定理

设f(x)=arctanx+arctan1/x,f(1)=arctan(1)+arctan(1)=π/2f'(x)=1/(1+x^2)+1/(1+(1/x)^2))*(-1/(x^2))=0对任意a>0,f(x)在[a,1](或[1,a])上连续,在(a,1)(或(1,a))上可导.根据中值定理：存在u,满足u在a与1之间,使得f'(...