Answer
$$y\left( x \right) = {x^2} + \ln x + 1$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{2{x^2} + 1}}{x},{\text{ }}y\left( 1 \right) = 2 \cr
& {\text{Applying the Formula }}\left( {11} \right){\text{ }}y\left( x \right) = {y_0} + \int_{{x_0}}^x {f\left( t \right)dt} \cr
& f\left( t \right) = \frac{{2{t^2} + 1}}{t},{\text{ }}{x_0} = 1,{\text{ }}{y_0} = 2 \cr
& {\text{Then,}} \cr
& y\left( x \right) = 2 + \int_1^x {\frac{{2{t^2} + 1}}{t}dt} \cr
& y\left( x \right) = 2 + \left[ {{t^2} + \ln t} \right]_1^x \cr
& y\left( x \right) = 2 + \left[ {{x^2} + \ln x} \right] - \left[ {{1^2} + \ln 1} \right] \cr
& y\left( x \right) = 2 + {x^2} + \ln x - 1 + \ln 1 \cr
& y\left( x \right) = {x^2} + \ln x + 1 \cr} $$
