The integral to evaluate is:

$\int \frac{x^3}{x^4 - 5} \, dx$

Solution:

1. Substitution: Let $u = x^4 - 5$. Then, $du = 4x^3 \, dx$.

2. Rewrite $x^3 \, dx$ in terms of $u$: $x^3 \, dx = \frac{1}{4} du$

3. Replace the integral in terms of $u$: $\int \frac{x^3}{x^4 - 5} \, dx = \int \frac{\frac{1}{4} du}{u}$

4. Simplify: $\frac{1}{4} \int \frac{1}{u} \, du$

5. Solve the integral: $\frac{1}{4} \ln|u| + C$

6. Substitute back $u = x^4 - 5$: $\frac{1}{4} \ln|x^4 - 5| + C$

Final Answer:

$\int \frac{x^3}{x^4 - 5} \, dx = \frac{1}{4} \ln|x^4 - 5| + C$

Let me know if you need further details or explanations.

Questions for Exploration:

1. What happens if the denominator is $x^4 + 5$ instead of $x^4 - 5$?
2. How can this substitution method be applied to trigonometric integrals?
3. What is the role of the absolute value in logarithmic solutions for integrals?
4. Can partial fraction decomposition be used for similar types of problems?
5. How does the substitution method change for definite integrals?

Tip:

Always check the domain of your substitution and logarithmic solutions to ensure they align with the original integral.