The problem asks for the integral of the function:

$\int \frac{x^2}{1 - x^6} \, dx$

Solution:

1. Substitution: Let $u = x^3$, so $du = 3x^2 \, dx$, or equivalently $\frac{1}{3} du = x^2 \, dx$.

2. Rewrite the integral: Substituting into the integral, we have: $\int \frac{x^2}{1 - x^6} \, dx = \int \frac{1}{3} \cdot \frac{du}{1 - u^2}$

3. Recognize the standard form: The integral is now in a standard form: $\int \frac{du}{1 - u^2} = \frac{1}{2} \ln \left|\frac{1 + u}{1 - u}\right| + C$ where $C$ is the constant of integration.

4. Substitute back $u = x^3$: $\frac{1}{3} \int \frac{du}{1 - u^2} = \frac{1}{6} \ln \left|\frac{1 + x^3}{1 - x^3}\right| + C$

Thus, the final answer is:

$\int \frac{x^2}{1 - x^6} \, dx = \frac{1}{6} \ln \left|\frac{1 + x^3}{1 - x^3}\right| + C$

Would you like further details on any step or have any questions?

Here are five questions to expand on this topic:

1. How can you determine when to use substitution in integration?
2. What are other methods to solve integrals of rational functions?
3. How does the form of the integrand help in recognizing standard integrals?
4. What is the significance of the absolute value in the logarithmic function?
5. Can partial fraction decomposition be applied to solve this integral? How?

Tip: When faced with an integral involving a complex fraction, substitution often simplifies the problem significantly.