Answer
$$y = x - 2{\tan ^{ - 1}}x + \pi - 1$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{{x^2} - 1}}{{{x^2} + 1}},\,\,\,\,y\left( 1 \right) = \frac{\pi }{2} \cr
& {\text{Using long division }}\frac{{{x^2} - 1}}{{{x^2} + 1}} = 1 - \frac{2}{{{x^2} + 1}} \cr
& \frac{{dy}}{{dx}} = 1 - \frac{2}{{{x^2} + 1}} \cr
& {\text{Separating variables}} \cr
& dy = \left( {1 - \frac{2}{{{x^2} + 1}}} \right)dx \cr
& {\text{Integrating both sides of the equation}} \cr
& \int {dy} = \int {\left( {1 - \frac{2}{{{x^2} + 1}}} \right)} dx \cr
& y = x - 2{\tan ^{ - 1}}x + C \cr
& \cr
& {\text{Using the initial condition }}y\left( 1 \right) = \frac{\pi }{2} \cr
& \frac{\pi }{2} = 1 - 2{\tan ^{ - 1}}\left( 1 \right) + C \cr
& \frac{\pi }{2} = 1 - 2\left( {\frac{\pi }{4}} \right) + C \cr
& \frac{\pi }{2} + \frac{\pi }{2} - 1 = C \cr
& C = \pi - 1 \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& y = x - 2{\tan ^{ - 1}}x + \pi - 1 \cr} $$
