The angle between the two curves $\mathrm{y}=\frac{{\mathrm{x}}^2-7\mathrm{x}+11}{\mathrm{x}-1}$ and $\mathrm{y}=\frac{\mathrm{x}+3}{{\mathrm{x}}^2+1}$at (2, 1) is 

$\begin{array}{rcl}\mathrm{y}& =& \frac{{\mathrm{x}}^2-7\mathrm{x}+11}{\mathrm{x}-1} \mathrm{at} (2, 1)\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{(\mathrm{x}-1) (2\mathrm{x}-7)-{({\mathrm{x}}^2-7\mathrm{x}+11)}^1}{{(\mathrm{x}-1)}^2}\\ {\mathrm{m}}_1& =& {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(2, 1)}\\ & =& \frac{1(-3)-1}{{(2-1)}^2}\\ {\mathrm{m}}_1& =& -4\\ \mathrm{y}& =& \frac{\mathrm{x}+3}{{\mathrm{x}}^2+1}\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{({\mathrm{x}}^2+1) \left(1\right)-(\mathrm{x}+3) \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2+1)}^2}\end{array}$

$\begin{array}{rcl}{\mathrm{m}}_2& =& {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(2, 1)}\\ & =& \frac{5-20}{{\left(5\right)}^2}=\frac{-15}{25}\\ {\mathrm{m}}_2& =& \frac{-3}{5}\\ \mathrm{tan\theta }& =& \left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1 {\mathrm{m}}_2}\right|\\ & =& \left|\frac{-4+{\displaystyle \frac{3}{5}}}{1+{\displaystyle \frac{12}{5}}}\right|\\ & =& \left|\frac{{\displaystyle \frac{17}{5}}}{{\displaystyle \frac{17}{5}}}\right|=1\\ \mathrm{\theta }& =& \frac{\mathrm{\pi }}{4}\end{array}$