To solve this problem, we need to find the integral of the function $(4x - 3)^4 \, dx$.

Step-by-Step Solution:

1. Identify the Substitution:

   Let $u = 4x - 3$. Then, differentiate with respect to $x$: $\frac{du}{dx} = 4 \implies dx = \frac{du}{4}$

2. Rewrite the Integral:

   Substituting $u = 4x - 3$ and $dx = \frac{du}{4}$, the integral becomes: $\int (4x - 3)^4 \, dx = \int u^4 \cdot \frac{du}{4} = \frac{1}{4} \int u^4 \, du$

3. Integrate with Respect to $u$:

   The integral of $u^4$ is: $\int u^4 \, du = \frac{u^5}{5} + C$

   Therefore, $\frac{1}{4} \int u^4 \, du = \frac{1}{4} \left( \frac{u^5}{5} \right) + C = \frac{1}{20} u^5 + C$

4. Substitute Back $u = 4x - 3$:

   Now substitute $u = 4x - 3$ back into the expression: $\frac{1}{20} (4x - 3)^5 + C$

5. Conclusion:

   The correct answer is: $\boxed{\text{d. } \frac{1}{20} (4x - 3)^5 + C}$

Would you like more details or have any questions?

Related Questions:

1. How do you perform integration using substitution in more complex cases?
2. What are some common mistakes to avoid when integrating polynomials?
3. How do you differentiate between indefinite and definite integrals?
4. What are the fundamental theorems of calculus, and how do they apply to integration?
5. Can you explain the relationship between integration and the area under a curve?

Tip: Always check your substitution and limits (if any) when solving integrals to ensure correctness.