1. **Problem statement:** Calculate the antiderivative (indefinite integral) of the function $$\int \frac{x}{\sqrt{1-x^4}} \, dx$$. 2. **Formula and approach:** To solve integrals involving expressions like $$\sqrt{1-x^4}$$, substitution is often useful. Here, notice that the denominator involves $$1 - x^4$$, which can be rewritten as $$1 - (x^2)^2$$. 3. **Substitution:** Let $$u = x^2$$, then $$du = 2x \, dx$$ or $$x \, dx = \frac{du}{2}$$. 4. **Rewrite the integral:** $$\int \frac{x}{\sqrt{1-x^4}} \, dx = \int \frac{x}{\sqrt{1-(x^2)^2}} \, dx = \int \frac{x}{\sqrt{1-u^2}} \, dx$$ Using substitution, replace $$x \, dx$$ with $$\frac{du}{2}$$: $$= \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{1}{\sqrt{1-u^2}} \, du$$ 5. **Integral formula:** Recall that $$\int \frac{1}{\sqrt{1-u^2}} \, du = \arcsin(u) + C$$ 6. **Apply the formula:** $$\frac{1}{2} \int \frac{1}{\sqrt{1-u^2}} \, du = \frac{1}{2} \arcsin(u) + C$$ 7. **Back-substitute:** Replace $$u = x^2$$: $$\int \frac{x}{\sqrt{1-x^4}} \, dx = \frac{1}{2} \arcsin(x^2) + C$$ **Final answer:** $$\boxed{\int \frac{x}{\sqrt{1-x^4}} \, dx = \frac{1}{2} \arcsin(x^2) + C}$$