The integral provided is:

$\int \left( (x+1)e^{x^2 + 2x + 1} + 6x^2 \sin(2x^3) \right) dx$

Let's solve this step by step.

Step 1: Simplify the expression

1. The term $e^{x^2 + 2x + 1}$ can be simplified as $e^{(x+1)^2}$, because $x^2 + 2x + 1 = (x+1)^2$.

   Thus, the first part of the integrand becomes: $(x+1)e^{(x+1)^2}$

2. The second term $6x^2 \sin(2x^3)$ remains as it is.

Step 2: Break the integral into two parts

$\int \left( (x+1)e^{(x+1)^2} + 6x^2 \sin(2x^3) \right) dx$

Can be written as:

$\int (x+1)e^{(x+1)^2} dx + \int 6x^2 \sin(2x^3) dx$

Step 3: Solve each integral

1. First integral: $\int (x+1)e^{(x+1)^2} dx$

   Let $u = (x+1)^2$, then $du = 2(x+1) dx$, or $\frac{du}{2} = (x+1) dx$.

   So, this becomes: $\int e^u \cdot \frac{du}{2} = \frac{1}{2} \int e^u du = \frac{1}{2} e^u = \frac{1}{2} e^{(x+1)^2}$

   Therefore: $\int (x+1)e^{(x+1)^2} dx = \frac{1}{2} e^{(x+1)^2}$

2. Second integral: $\int 6x^2 \sin(2x^3) dx$

   Let $v = 2x^3$, then $dv = 6x^2 dx$.

   So, this becomes: $\int 6x^2 \sin(2x^3) dx = \int \sin(v) dv = -\cos(v) = -\cos(2x^3)$

Step 4: Combine the results

Now, adding the results of both integrals, we get:

$\frac{1}{2} e^{(x+1)^2} - \cos(2x^3) + C$

Where $C$ is the constant of integration.

Final Answer:

$\int \left( (x+1)e^{x^2 + 2x + 1} + 6x^2 \sin(2x^3) \right) dx = \frac{1}{2} e^{(x+1)^2} - \cos(2x^3) + C$

Would you like any further details or have questions about this solution?

5 Related Questions:

1. How does substitution work in integration?
2. Can this method be used for similar exponential integrals?
3. What are the common techniques for solving trigonometric integrals?
4. How do we recognize when to apply substitution in integrals?
5. What is the general process for solving definite integrals?

Tip:

Always look for patterns like perfect squares in the exponent of exponential functions or powers in trigonometric functions to simplify complex integrals.