Let's solve the given integral:

Problem 29: $\int (5x^2 + 1)(5x^3 + 3x - 8)^6 \, dx$

This problem requires the method of substitution. The goal is to simplify the integral by substituting a part of the integrand.

1. Substitution: Let's choose $u = 5x^3 + 3x - 8$. Now differentiate $u$ with respect to $x$: $\frac{du}{dx} = 15x^2 + 3 \quad \text{or} \quad du = (15x^2 + 3) \, dx$ Notice that $15x^2 + 3$ can be factored as $3(5x^2 + 1)$. Thus, $du = 3(5x^2 + 1) \, dx \quad \text{or} \quad \frac{du}{3} = (5x^2 + 1) \, dx$

2. Rewrite the integral: Substitute $u = 5x^3 + 3x - 8$ into the integral: $\int (5x^2 + 1)(5x^3 + 3x - 8)^6 \, dx = \int (5x^3 + 3x - 8)^6 \cdot \frac{du}{3}$ This simplifies to: $\frac{1}{3} \int u^6 \, du$

3. Integrate: Now integrate $u^6$: $\int u^6 \, du = \frac{u^7}{7} + C$ So, the integral becomes: $\frac{1}{3} \cdot \frac{u^7}{7} + C = \frac{1}{21} u^7 + C$

4. Substitute back: Replace $u$ with $5x^3 + 3x - 8$: $\int (5x^2 + 1)(5x^3 + 3x - 8)^6 \, dx = \frac{1}{21} (5x^3 + 3x - 8)^7 + C$

Final Answer:

$\int (5x^2 + 1)(5x^3 + 3x - 8)^6 \, dx = \frac{1}{21} (5x^3 + 3x - 8)^7 + C$

Would you like more details or have any other questions?

Related Questions

1. How would you approach this integral if the exponent was different, such as $(5x^3 + 3x - 8)^n$?
2. Can this method be used for integrals involving trigonometric functions like $\int (1 + \sin(x))(\cos(x))^5 \, dx$?
3. What happens if the integrand includes a product of more than two polynomials?
4. How do you evaluate the integral if $u$ substitution doesn't simplify the expression easily?
5. Can you use integration by parts instead of substitution for this problem?

Tip:

When choosing a substitution, look for expressions inside the integrand whose derivative also appears elsewhere in the integrand. This simplifies the process significantly.