To solve the integral $\int \frac{x^3}{\sqrt{1 - x^4}} \, dx$ using $u$-substitution, we can follow these steps:

1. Identify the substitution: Let's choose $u$ such that the integrand simplifies significantly. A good candidate is $u = 1 - x^4$, since its derivative, when multiplied by $x^3$, will help us simplify the integral.

2. Compute the derivative: We need to find $du$ in terms of $dx$. $u = 1 - x^4 \implies \frac{du}{dx} = -4x^3 \implies du = -4x^3 \, dx$ Solving for $dx$, we get: $dx = \frac{du}{-4x^3}$

3. Substitute $u$ and $dx$ in the integral: Substitute $u$ and $dx$ into the integral: $\int \frac{x^3}{\sqrt{1 - x^4}} \, dx = \int \frac{x^3}{\sqrt{u}} \cdot \frac{du}{-4x^3}$

4. Simplify the integral: $\int \frac{x^3}{\sqrt{u}} \cdot \frac{du}{-4x^3} = \int \frac{1}{-4\sqrt{u}} \, du = -\frac{1}{4} \int u^{-\frac{1}{2}} \, du$

5. Integrate with respect to $u$: $-\frac{1}{4} \int u^{-\frac{1}{2}} \, du = -\frac{1}{4} \cdot 2u^{\frac{1}{2}} + C = -\frac{1}{2} u^{\frac{1}{2}} + C$

6. Substitute back $u = 1 - x^4$: $-\frac{1}{2} u^{\frac{1}{2}} + C = -\frac{1}{2} \sqrt{1 - x^4} + C$

Therefore, the solution to the integral $\int \frac{x^3}{\sqrt{1 - x^4}} \, dx$ is: $-\frac{1}{2} \sqrt{1 - x^4} + C$

Would you like any further details or have any questions?

Here are 8 related questions to further your understanding:

1. How would the integral change if the integrand was $\int \frac{x^2}{\sqrt{1 - x^4}} \, dx$?
2. Can you use a different substitution to solve the same integral?
3. What if the integral was $\int \frac{x^3}{\sqrt{1 + x^4}} \, dx$?
4. How do you determine the best substitution for an integral?
5. Can this integral be solved using integration by parts?
6. How does the constant of integration $C$ affect the final solution?
7. What are some common substitutions for integrals involving square roots?
8. How does $u$-substitution simplify integrals involving polynomial expressions?

Tip: Always double-check your substitution and the corresponding derivative to ensure you are integrating correctly.