Let’s solve the problems step by step.

1. Solve $\int (5x - 3)^4 \, dx$

We use substitution: Let $u = 5x - 3$, then $\frac{du}{dx} = 5 \Rightarrow dx = \frac{du}{5}$.

The integral becomes: $\int (5x - 3)^4 dx = \int u^4 \cdot \frac{1}{5} du = \frac{1}{5} \int u^4 du$

The integral of $u^4$ is: $\frac{u^5}{5} + C = \frac{(5x - 3)^5}{25} + C$

Answer: $\frac{(5x - 3)^5}{25} + C$

2. Solve $\int 4x \cdot (6x^2 - 5)^4 \, dx$

Again, use substitution: Let $u = 6x^2 - 5$, so $\frac{du}{dx} = 12x \Rightarrow dx = \frac{du}{12x}$.

The $x$ in the numerator cancels with the $x$ in the denominator: $\int 4x (6x^2 - 5)^4 dx = \int \frac{4}{12} u^4 du = \frac{1}{3} \int u^4 du$

The integral of $u^4$ is: $\frac{u^5}{5} + C = \frac{1}{3} \cdot \frac{(6x^2 - 5)^5}{5} + C$

Simplify: $\frac{(6x^2 - 5)^5}{15} + C$

Answer: $\frac{(6x^2 - 5)^5}{15} + C$

3. Solve $\int_0^1 (8x^3 + 6x^2 - 2x) \, dx$

We integrate term by term: $\int_0^1 8x^3 dx = \left[2x^4\right]_0^1 = 2(1)^4 - 2(0)^4 = 2$ $\int_0^1 6x^2 dx = \left[2x^3\right]_0^1 = 2(1)^3 - 2(0)^3 = 2$ $\int_0^1 -2x dx = \left[-x^2\right]_0^1 = -(1)^2 - -(0)^2 = -1$

Add them together: $2 + 2 - 1 = 3$

Answer: $3$

Let me know if you want further clarifications! Here are related questions you can consider:

1. What are the steps for solving definite integrals using substitution?
2. How do polynomial integrals work term-by-term?
3. How do you verify if substitution is the correct method?
4. How do definite integrals differ from indefinite integrals in terms of constants?
5. What happens when a substitution introduces a new integration bound?

Tip: Always check your substitution carefully to