Answer
$$\displaystyle{\int x\sqrt{1-x^4}dx=\frac{x^2 \sqrt{1-x^4}}{4}+\frac{1}{4}\arcsin\left( x^2\right)+C}$$
Work Step by Step
$\displaystyle{I=\int x\sqrt{1-x^4}dx}\\ \displaystyle{I=\int x\sqrt{1-\left(x^2\right)^2}dx}\\ $
$\displaystyle \left[\begin{array}{ll} x^2=\sin\theta & x^4=\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{\cos\theta}{2\sqrt{\sin\theta}} & dx=\frac{\cos\theta}{2\sqrt{\sin\theta}}\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int \sqrt{\sin\theta}\sqrt{1-\sin^2\theta}\times\frac{\cos\theta}{2\sqrt{\sin\theta}}\ d\theta}\\ \displaystyle{I=\int \frac{1}{2}\cos^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{2}\int \cos^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{4}\int \cos2\theta+1\ d\theta}\\ \displaystyle{I=\frac{1}{4}\left(\frac{1}{2} \sin2\theta+\theta\right)+C}\\ \displaystyle{I=\frac{1}{8}\sin2\theta+\frac{1}{4}\theta+C}\\ \displaystyle{I=\frac{1}{4}\sin\theta\cos\theta+\frac{1}{4}\theta+C}\\$
$\sin\theta=x^2\\ \cos\theta=\sqrt{1-x^4}\\ $
$\displaystyle{I=\frac{1}{4}\times x^2\times \sqrt{1-x^4}+\frac{1}{4}\arcsin\left( x^2\right)+C}\\ \displaystyle{I=\frac{x^2 \sqrt{1-x^4}}{4}+\frac{1}{4}\arcsin\left( x^2\right)+C}\\ $
