Let $g(x)=f(x)+f(\frac{1}{x})$. Then we have

$$g'(x) = f'(x)-\frac{1}{x^2}f'(x)=\frac{\ln{x}}{1+x}-\frac{1}{x^2}\frac{\ln\frac{1}{x}}{1+\left(\frac{1}{x}\right)^2}=\frac{\ln{x}}{x}$$

Therefore

$$g(x)=\int \frac{\ln{x}}{x}dx=\int \ln{x}d\ln{x}=\frac{(\ln{x})^2}{2}+C$$

By the conclusion of # 4543, we have $g(2)=\frac{(\ln{x})^2}{2}$. This means the constant $C=0$. Hence the result is $$\boxed{\frac{(\ln{x})^2}{2}}$$