Answer
$$y = \frac{x}{{\sqrt {{x^2} + 1} }} + 1$$
Work Step by Step
$$\eqalign{
& {\left( {{x^2} + 1} \right)^2}\frac{{dy}}{{dx}} = \sqrt {{x^2} + 1} ,\,\,\,\,\,\,\,y\left( 0 \right) = 1 \cr
& \cr
& {\text{separate the variables}} \cr
& \frac{{dy}}{{dx}} = \frac{{\sqrt {{x^2} + 1} }}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr
& \cr
& {\text{Use Trigonometric Substitutions:}} \cr
& {\text{We set}}{\text{, }}x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& y = \int {\frac{1}{{{{\left( {{{\tan }^2}\theta + 1} \right)}^{3/2}}}}} \left( {{{\sec }^2}\theta d\theta } \right) \cr
& y = \int {\frac{1}{{{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}} \left( {{{\sec }^2}\theta d\theta } \right) \cr
& y = \int {\frac{1}{{{{\sec }^3}\theta }}} \left( {{{\sec }^2}\theta d\theta } \right) \cr
& y = \int {\frac{1}{{\sec \theta }}} d\theta \cr
& y = \int {\cos \theta } d\theta \cr
& y = \sin \theta + C \cr
& \cr
& {\text{We have that }}x = \tan \theta {\text{; then sin}}\theta = \frac{x}{{\sqrt {{x^2} + 1} }} \cr
& y = \frac{x}{{\sqrt {{x^2} + 1} }} + C\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{use initial condition }}y\left( 0 \right) = 1 \cr
& 1 = \frac{0}{{\sqrt {{0^2} + 1} }} + C\,\, \cr
& C = 1 \cr
& \cr
& {\text{Then}}{\text{, substituting }}C = 1{\text{ in }}\left( {\bf{1}} \right) \cr
& y = \frac{x}{{\sqrt {{x^2} + 1} }} + 1 \cr} $$
