Let $\mathrm{f}:\mathrm{R}\to [4,\mathrm{\infty })$ be an onto quadratic function whose leading coefficient is 1, such that $\mathrm{f}\text{'}\left(\mathrm{x}\right)+\mathrm{f}\text{'}(2-\mathrm{x})=0$. Then the value of  ${\int }_1^3 \frac{\mathrm{dx}}{\mathrm{f}\left(\mathrm{x}\right)}$ is equal to:

$\begin{array}{l}\mathrm{f}\text{'}\left(\mathrm{x}\right)+\mathrm{f}\text{'}(2-\mathrm{x})=0\\ \mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}(2-\mathrm{x})=\mathrm{\lambda }\end{array}$

Put $\mathrm{x}=1$, we get $\mathrm{\lambda }=0$

$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(2-\mathrm{x})$

Because f is quadratic function therefore $\mathrm{f}\left(\mathrm{x}\right)$ is symmetrical about line $\mathrm{x}=1$

$\begin{array}{l}\therefore \mathrm{f}\left(1\right)=4\\ \therefore \mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)=(\mathrm{x}-1{)}^2+4\\ \therefore {\int }_1^3 \frac{\mathrm{dx}}{(\mathrm{x}-1{)}^2+4}={\left[\frac{1}{2}{\tan }^{-1}\frac{\mathrm{x}-1}{2}\right]}_1^3=\frac{1}{2}\left[{\tan }^{-1}1-{\tan }^{-1}\mathrm{O}\right]=\frac{\mathrm{\pi }}{8}\end{array}$