Answer
$$y = \tan \left( {x + \frac{\pi }{4}} \right)$$
Work Step by Step
$$\eqalign{
& y' = 1 + {y^2},\,\,\,\,y\left( 0 \right) = 1 \cr
& \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = 1 + {y^2} \cr
& {\text{separating the variables}} \cr
& \frac{{dy}}{{1 + {y^2}}} = dx \cr
& \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {\frac{{dy}}{{1 + {y^2}}}} = \int {dx} \cr
& {\tan ^{ - 1}}y = x + C \cr
& \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 1 \cr
& {\tan ^{ - 1}}1 = 0 + C \cr
& C = \frac{\pi }{4} \cr
& \cr
& {\text{Then}}{\text{, the particular solution is}} \cr
& {\tan ^{ - 1}}y = x + \frac{\pi }{4} \cr
& {\text{solve for }}y \cr
& \tan \left( {{{\tan }^{ - 1}}y} \right) = \tan \left( {x + \frac{\pi }{4}} \right) \cr
& y = \tan \left( {x + \frac{\pi }{4}} \right) \cr} $$
