令f(x)=arctan x+acrtan (1／x)则f'(x) = 1/(1+x²) + 1/(1+(1/x)²) * (-1/x²)= 1/(1+x²) - 1/(1+x²)=0就是说f(x)=c是一个常数令x=1有,f(1)=arctan1 + arctan1 =π／4 +π／4 =π／2 =c所以...