Answer
$$ - \frac{1}{4}{\operatorname{csch} ^{ - 1}}\left| x \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {1 + {x^4}} }}} \cr
& {\text{rewriting }} \cr
& = \int {\frac{{2xdx}}{{2{x^2}\sqrt {{{\left( 1 \right)}^2} + {{\left( {{x^2}} \right)}^2}} }}} \cr
& = \frac{1}{2}\int {\frac{{2xdx}}{{{x^2}\sqrt {{{\left( 1 \right)}^2} + {{\left( {{x^2}} \right)}^2}} }}} \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.1}}2 \cr
& \int {\frac{{dx}}{{u\sqrt {{a^2} + {x^2}} }}} = - \frac{1}{a}{\operatorname{csch} ^{ - 1}}\frac{{\left| u \right|}}{a} + C \cr
& ,so \cr
& = \frac{1}{4}\left( { - \frac{1}{1}{{\operatorname{csch} }^{ - 1}}\frac{{\left| x \right|}}{1}} \right) + C \cr
& {\text{simplify}} \cr
& = - \frac{1}{4}{\operatorname{csch} ^{ - 1}}\left| x \right| + C \cr} $$
