The integral you're asking for is quite complex. Here's a step-by-step solution:

1. First, let's simplify the expression. We can rewrite the integral as ∫6x(9x^2 + 4)^4 dx.

2. Now, let's use a substitution to simplify the integral. Let u = 9x^2 + 4. Then, du = 18x dx.

3. We can see that we have x dx in our integral, but we need 18x dx for our substitution. So, we can rewrite the integral as 1/3 ∫18x(9x^2 + 4)^4 dx.

4. Now, we can substitute u into the integral: 1/3 ∫u^4 du.

5. The integral of u^4 is 1/5 u^5, so we have 1/3 * 1/5 u^5 = 1/15 u^5.

6. Finally, we substitute u back in for 9x^2 + 4 to get our final answer: 1/15 (9x^2 + 4)^5 + C, where C is the constant of integration.

So, the integral of 6x(3x^2 + 4)^4 dx is 1/15 (9x^2 + 4)^5 + C.

Question

The integral you're asking for is quite complex. Here's a step-by-step solution:

1. First, let's simplify the expression. We can rewrite the integral as ∫6x(9x^2 + 4)^4 dx.

2. Now, let's use a substitution to simplify the integral. Let u = 9x^2 + 4. Then, du = 18x dx.

3. We can see that we have x dx in our integral, but we need 18x dx for our substitution. So, we can rewrite the integral as 1/3 ∫18x(9x^2 + 4)^4 dx.

4. Now, we can substitute u into the integral: 1/3 ∫u^4 du.

5. The integral of u^4 is 1/5 u^5, so we have 1/3 * 1/5 u^5 = 1/15 u^5.

6. Finally, we substitute u back in for 9x^2 + 4 to get our final answer: 1/15 (9x^2 + 4)^5 + C, where C is the constant of integration.

So, the integral of 6x(3x^2 + 4)^4 dx is 1/15 (9x^2 + 4)^5 + C.

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