Answer
$$\frac{1}{4}{\sin ^{ - 1}}{x^4} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}dx} \cr
& {\text{Rewrte the integrand}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}dx} = \int {\frac{{{x^3}}}{{\sqrt {1 - {{\left( {{x^4}} \right)}^2}} }}dx} \cr
& {\text{Let }}u = {x^4} \Rightarrow du = 4{x^3}dx,{\text{ }}dx = \frac{{du}}{{4{x^3}}} \cr
& {\text{Applying the substitution}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {1 - {{\left( {{x^4}} \right)}^2}} }}dx} = \int {\frac{{{x^3}}}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{4{x^3}}}} \cr
& {\text{ }} = \frac{1}{4}\int {\frac{1}{{\sqrt {1 - {u^2}} }}du} \cr
& {\text{Integrating}} \cr
& \frac{1}{4}\int {\frac{1}{{\sqrt {1 - {u^2}} }}du} = \frac{1}{4}{\sin ^{ - 1}}u + C \cr
& {\text{Back - substitute }}u = {x^4} \cr
& {\text{ }} = \frac{1}{4}{\sin ^{ - 1}}{x^4} + C \cr} $$
