Answer
$$ y = {\tan ^{ - 1}}x - x + 1$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{{x^2} + 1}} - 1,\,\,\,\,\,y\left( 0 \right) = 1 \cr
& {\text{separating variables, we get:}} \cr
& dy = \left( {\frac{1}{{{x^2} + 1}} - 1} \right)dx \cr
& {\text{integrate both sides}} \cr
& y = \int {\left( {\frac{1}{{{x^2} + 1}} - 1} \right)} dx \cr
& y = {\tan ^{ - 1}}x - x + C\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{use initial condition }}y\left( 0 \right) = 1 \cr
& 1 = {\tan ^{ - 1}}0 - 0 + C \cr
& C = 1 \cr
& \cr
& {\text{then}}{\text{, substitute }}C = 1{\text{ in }}\left( {\bf{1}} \right) \cr
& y = {\tan ^{ - 1}}x - x + 1 \cr} $$
