The method in the image uses substitution and power rules for integration. Let's apply the same steps to solve the integral $\int x^2(2x^3 + 1)^4 \, dx$:

Step 1: Substitution

Let’s choose a substitution for $u$:

• Define $u = 2x^3 + 1$.
• Then, differentiate $u$: $\frac{du}{dx} = 6x^2 \quad \text{or} \quad du = 6x^2 \, dx$ This matches the $x^2 \, dx$ term in the original integral, so we can rewrite: $x^2 \, dx = \frac{1}{6} \, du$

Step 2: Rewrite the Integral

Now substitute into the original integral: $\int x^2(2x^3 + 1)^4 \, dx = \int (u^4) \cdot \frac{1}{6} \, du = \frac{1}{6} \int u^4 \, du$

Step 3: Integrate

Use the power rule for integration: $\int u^4 \, du = \frac{u^5}{5} + C$ Thus, the integral becomes: $\frac{1}{6} \cdot \frac{u^5}{5} + C = \frac{1}{30} u^5 + C$

Step 4: Substitute Back

Substitute back $u = 2x^3 + 1$: $\frac{1}{30} (2x^3 + 1)^5 + C$

Final Answer:

$\int x^2(2x^3 + 1)^4 \, dx = \frac{1}{30} (2x^3 + 1)^5 + C$

Do you have any questions or need further clarification?

Relative Questions:

1. How does substitution work in integration?
2. Can we use this method for trigonometric integrals?
3. What happens if the exponent is negative or a fraction in substitution?
4. How do we handle more complex expressions for $u$?
5. Could integration by parts be used here instead?

Tip:

Always check the differential after substitution to ensure the new integrand simplifies correctly.