f(x) + g(x) = 1/(x+1)
将此式中的x用-x代替，并利用奇偶性的条件，有
f(x) - g(x) = 1/(-x+1)
这是个二元一次方程组
f(x) = 1/2[1/(x+1) + 1/(-x+1)] = 1/(1-x^2)
g(x) = 1/2[1/(x+1) - 1/(-x+1)] = x/(x^2-1)
f(x) + g(x) = 1/(x+1)
将此式中的x用-x代替，并利用奇偶性的条件，有
f(x) - g(x) = 1/(-x+1)
这是个二元一次方程组
f(x) = 1/2[1/(x+1) + 1/(-x+1)] = 1/(1-x^2)
g(x) = 1/2[1/(x+1) - 1/(-x+1)] = x/(x^2-1)