Answer
$$A = \frac{\pi }{8}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{x^2} - 2x + 5}} \cr
& {\text{The area of the region is given by}} \cr
& A = \int_1^3 {\frac{1}{{{x^2} - 2x + 5}}} dx \cr
& {\text{Completing the square}} \cr
& A = \int_1^3 {\frac{1}{{{x^2} - 2x + 1 + 4}}} dx \cr
& A = \int_1^3 {\frac{1}{{{{\left( {x - 1} \right)}^2} + 4}}} dx \cr
& {\text{Let }}u = x - 1,{\text{ }}du = dx \cr
& {\text{The new limits of integration are}} \cr
& x = 3,{\text{ }}u = 3 - 1 = 2 \cr
& x = 1,{\text{ }}u = 1 - 1 = 0 \cr
& {\text{Substituting}} \cr
& A = \int_0^2 {\frac{1}{{{u^2} + 4}}} dx \cr
& {\text{Integrate using basic integration rules}} \cr
& A = \left[ {\frac{1}{2}{{\tan }^{ - 1}}\frac{u}{2}} \right]_0^2 \cr
& A = \frac{1}{2}\left( {{{\tan }^{ - 1}}\frac{2}{2} - {{\tan }^{ - 1}}\frac{0}{2}} \right) \cr
& {\text{Simplify}} \cr
& A = \frac{1}{2}\left( {\frac{\pi }{4}} \right) \cr
& A = \frac{\pi }{8} \cr} $$
