1. **State the problem:** We want to solve the integral $$\int 5x \sqrt{1 - x^4} \, dx$$. 2. **Identify a substitution:** Notice the expression inside the square root is $1 - x^4$. Let’s set $$u = 1 - x^4$$. 3. **Differentiate $u$:** $$\frac{du}{dx} = -4x^3$$, so $$du = -4x^3 \, dx$$. 4. **Rewrite the integral:** We have $5x \sqrt{1 - x^4} \, dx = 5x \sqrt{u} \, dx$. But $du$ involves $x^3 dx$, and we only have $x dx$ in the integral, so direct substitution is not straightforward. 5. **Rewrite $x^4$ as $(x^2)^2$ and try substitution:** Let’s try $t = x^2$, then $dt = 2x dx$, so $x dx = \frac{dt}{2}$. 6. **Rewrite the integral in terms of $t$:** $$\int 5x \sqrt{1 - x^4} \, dx = \int 5 \sqrt{1 - t^2} \cdot x dx = \int 5 \sqrt{1 - t^2} \cdot \frac{dt}{2} = \frac{5}{2} \int \sqrt{1 - t^2} \, dt$$ 7. **Integral formula:** The integral $$\int \sqrt{1 - t^2} \, dt$$ is a standard form with solution: $$\frac{t}{2} \sqrt{1 - t^2} + \frac{1}{2} \arcsin(t) + C$$ 8. **Apply the formula:** $$\frac{5}{2} \int \sqrt{1 - t^2} \, dt = \frac{5}{2} \left( \frac{t}{2} \sqrt{1 - t^2} + \frac{1}{2} \arcsin(t) \right) + C = \frac{5}{4} t \sqrt{1 - t^2} + \frac{5}{4} \arcsin(t) + C$$ 9. **Substitute back $t = x^2$:** $$\boxed{\frac{5}{4} x^2 \sqrt{1 - x^4} + \frac{5}{4} \arcsin(x^2) + C}$$ This is the solution to the integral.