Cho hàm số $f\left( x \right)$ liên tục và nhận giá trị dương trên...

Ta có: $1+f\left( x \right)=f\left( x \right)f\left( 1-x \right)+f\left( x \right)\Rightarrow \dfrac{f\left( x \right)}{1+f\left( x \right)}=\dfrac{1}{f\left( 1-x \right)+1}$
Xét $I=\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( x \right)}}$
Đặt $t=1-x\Rightarrow x=1-t\Rightarrow dx=-dt.$
Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=0 \\
\end{aligned} \right..$
Khi đó $I=-\int\limits_{1}^{0}{\dfrac{dt}{1+f\left( 1-t \right)}=\int\limits_{0}^{1}{\dfrac{dt}{1+f\left( 1-t \right)}=\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( 1-x \right)}=\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+f\left( x \right)}dx.}}}}$
Mặt khác $\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( x \right)}+\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+f\left( x \right)}dx}=\int\limits_{0}^{1}{\dfrac{1+f\left( x \right)}{1+f\left( x \right)}dx}=\int\limits_{0}^{1}{dx}=1}$ hay $2I=1.$
Vậy $I=\dfrac{1}{2}.$