Answer
$$\frac{1}{2}\ln \left| {{x^2} + \sqrt {1 + {x^4}} } \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{\sqrt {1 + {x^4}} }}} \cr
& = \int {\frac{{xdx}}{{\sqrt {1 + {{\left( {{x^2}} \right)}^2}} }}} \cr
& {\text{Set }}y = {x^2},\,\,\,\,dy = 2xdx \cr
& \int {\frac{{xdx}}{{\sqrt {1 + {{\left( {{x^2}} \right)}^2}} }}} = \frac{1}{2}\int {\frac{{dy}}{{\sqrt {1 + {y^2}} }}} \cr
& {\text{Using Trigonometric Substitutions,}} \cr
& {\text{we set}}{\text{, }}y = \tan \theta,{\text{ }}dy = {\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \frac{1}{2}\int {\frac{{dy}}{{\sqrt {1 + {y^2}} }}} = \frac{1}{2}\int {\frac{{{{\sec }^2}\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}} d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \frac{1}{2}\int {\frac{{{{\sec }^2}\theta }}{{\sqrt {{{\sec }^2}\theta } }}} d\theta \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\int {\sec \theta d\theta } \cr
& {\text{Integrate}} \cr
& = \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{write in terms of }}y,{\text{ }}\tan \theta = y{\text{ and sec}}\theta = \sqrt {1 + {y^2}} \cr
& = \frac{1}{2}\ln \left| {y + \sqrt {1 + {y^2}} } \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}y = {x^2} \cr
& = \frac{1}{2}\ln \left| {{x^2} + \sqrt {1 + {x^4}} } \right| + C \cr} $$
