Answer
$$\frac{1}{2}{\sin ^{ - 1}}{x^2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {1 - {x^4}} }}} dx \cr
& {\text{substitute }}u = {x^2},{\text{ }}du = 2xdx \cr
& \int {\frac{x}{{\sqrt {1 - {x^4}} }}} dx = \int {\frac{{\left( {1/2} \right)du}}{{\sqrt {1 - {u^2}} }}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{find the antiderivative using }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\frac{u}{a} + C{\text{ }}\left( {{\text{see page 468}}} \right) \cr
& = \frac{1}{2}{\sin ^{ - 1}}u + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {x^2} \cr
& = \frac{1}{2}{\sin ^{ - 1}}{x^2} + C \cr} $$
