1. **State the problem:** We need to evaluate the integral $$\int 7x \sqrt{1 - x^4} \, dx$$. 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 $7x \sqrt{1 - x^4} dx = 7x \sqrt{u} dx$. But $du$ involves $x^3 dx$, and we only have $x dx$ in the integral, so direct substitution is not straightforward. 5. **Rewrite the integral to match substitution:** Express $x$ terms to match $du$. Note that $x^4 = (x^2)^2$, so try substitution $t = x^2$, then $dt = 2x dx$ or $x dx = \frac{dt}{2}$. 6. **Rewrite integral in terms of $t$:** $$\int 7x \sqrt{1 - x^4} dx = 7 \int x \sqrt{1 - (x^2)^2} dx = 7 \int \sqrt{1 - t^2} \cdot x dx = 7 \int \sqrt{1 - t^2} \cdot \frac{dt}{2} = \frac{7}{2} \int \sqrt{1 - t^2} dt$$ 7. **Integral formula:** $$\int \sqrt{1 - t^2} dt = \frac{t}{2} \sqrt{1 - t^2} + \frac{1}{2} \arcsin(t) + C$$ 8. **Apply formula:** $$\frac{7}{2} \int \sqrt{1 - t^2} dt = \frac{7}{2} \left( \frac{t}{2} \sqrt{1 - t^2} + \frac{1}{2} \arcsin(t) \right) + C = \frac{7t}{4} \sqrt{1 - t^2} + \frac{7}{4} \arcsin(t) + C$$ 9. **Back-substitute $t = x^2$:** $$\boxed{\frac{7x^2}{4} \sqrt{1 - x^4} + \frac{7}{4} \arcsin(x^2) + C}$$ This is the evaluated integral.