Answer
$$ - \frac{1}{8}{\operatorname{sech} ^{ - 1}}\frac{{{x^4}}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {4 - {x^8}} }}} \cr
& {\text{rewriting }} \cr
& = \int {\frac{{4{x^3}dx}}{{4{x^4}\sqrt {{{\left( 2 \right)}^2} - {{\left( {{x^4}} \right)}^2}} }}} \cr
& = \frac{1}{4}\int {\frac{{4{x^3}dx}}{{{x^4}\sqrt {{{\left( 2 \right)}^2} - {{\left( {{x^4}} \right)}^2}} }}} \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.1}}2 \cr
& \int {\frac{{dx}}{{u\sqrt {{a^2} - {u^2}} }}} = - \frac{1}{a}{\operatorname{sech} ^{ - 1}}\frac{u}{a} + C \cr
& ,so \cr
& = \frac{1}{4}\left( { - \frac{1}{2}{{\operatorname{sech} }^{ - 1}}\frac{{{x^4}}}{2}} \right) + C \cr
& {\text{simplify}} \cr
& = - \frac{1}{8}{\operatorname{sech} ^{ - 1}}\frac{{{x^4}}}{2} + C \cr} $$
