f(x) is differentiable function satisfying f(x).f’(-x)=f(-x).f’(x), and f(0)=1, find value of integration of 1/{1+f(x)} from -5 to 5

if f(x) is even function, integration of f(x) from -5 to 5 = 2. integration of f(x) from 0 to 5, so how I get 0? if f(x) is an odd function, the result will be 0.

let f(x)=x^2 -1. 1/1+f(x) =1/x^2 is even also. now integrating from -5 to 5 how I get 0? plz clarify. 

sorry. but take g(x)=1/1+f(x)

g(-x)=g(x). so g(x) is even. 

now integration of g(x) from-5 to5 = 2*integration of g(x) from 0 to 5