Integral over recurrence relationship

Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form $$ F_n=\frac{a_n+b_nf}{a_n+b_n+a_nf} $$ for $n\ge 1$, where $a_n$ and $b_n$ are constants. By the recurrence relation, these constants must follow $$ \left( \begin{array}{c} a_{n+1}\\ b_{n+1} \end{array} \right)=\left( \begin{array}{cc} 1&1\\ 1&0 \end{array} \right)\left( \begin{array}{c} a_n\\ b_n \end{array} \right), $$ with $$ \left( \begin{array}{c} a_1\\ b_1 \end{array} \right)=\left( \begin{array}{c} 1\\ 0 \end{array} \right). $$ Thanks to linear algebra, you may solve this Fibonacci-like $a_n$ and $b_n$ immediately.

Hope this could be somewhat helpful for you.