5. Given : \( \int 6 x^{3} e^{x^{2}+3} d x \). Using integration by parts \( \left(\int u d v=u v-\int v d u\right) \) and letting \( u=f(x) \), \( d v=3 g(x) d x \), find \( h(x) \) if \( \int v d u=h(x)+C \)
a.) \( 6 e^{x^{2}+3} \)
b.) \( 3 e^{-3} e^{x^{2}} \)
c.) \( \frac{3 e^{x^{2}+6}}{e^{3}} \)
d.) \( 2 e^{x^{2}+3} \)
Solve the above problem.

To solve the integral \( \int 6 x^{3} e^{x^{2}+3} \, dx \) using integration by parts, we need to follow these steps:

1. Identify \( u \) and \( dv \) such that \( du \) and \( v \) are easily computable.

Let's choose:
\[ u = x^2 \quad \text{and} \quad dv = 6x e^{x^2 + 3} \, dx \]

2. Compute \( du \) and \( v \):
\[ du = 2x \, dx \]
To find \( v \), we integrate \( dv \):
\[ v = \int 6x e^{x^2 + 3} \, dx \]

To integrate \( 6x e^{x^2 + 3} \, dx \), notice that:
\[ \int 6x e^{x^2 + 3} \, dx \]
Let \( w = x^2 + 3 \), then \( dw = 2x \, dx \) and \( x \, dx = \frac{1}{2} dw \).
Thus:
\[ \int 6x e^{x^2 + 3} \, dx = 6 \int e^{w} \frac{1}{2} dw = 3 \int e^{w} \, dw = 3e^w + C = 3e^{x^2 + 3} + C \]
So:
\[ v = 3e^{x^2 + 3} \]

3. Apply integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
\[ \int x^2 \cdot 6x e^{x^2 + 3} \, dx = x^2 \cdot 3e^{x^2 + 3} - \int 3e^{x^2 + 3} \cdot 2x \, dx \]
\[ \int 6x^3 e^{x^2 + 3} \, dx = 3x^2 e^{x^2 + 3} - \int 6x e^{x^2 + 3} \, dx \]

To find \( h(x) \), we need to integrate:
\[ \int 6x e^{x^2 + 3} \, dx = 3e^{x^2 + 3} \]

Thus:
\[ h(x) = 3e^{x^2 + 3} \]

Comparing the given choices:
a.) \( 6 e^{x^{2}+3} \)
b.) \( 3 e^{-3} e^{x^{2}} \)
c.) \( \frac{3 e^{x^{2}+6}}{e^{3}} \)
d.) \( 2 e^{x^{2}+3} \)

The correct answer is:
d.) \( 2 e^{x^{2}+3} \)

However, this choice does not seem to match our computation. It seems there might be a mistake or misunderstanding in the given choices. Let's double-check the correct calculation steps:
\[ \int 6 x^{3} e^{x^{2}+3} \, dx \]
Here, correctly integrating:
\[ u = x^3, \quad dv = 6e^{x^2+3} dx \]
\[ du = 3x^2 \, dx, \quad v = 3e^{x^2+3} \]

Finally:
\[ \int v \, du = \int 3e^{x^2+3} \cdot 2x \, dx = \frac{3 e^{x^2+6}}{e^{3}} \]
Then \( h(x) \) would be:
\[ h(x) = 2 e^{x^{2}+3} \]

Reconsidering, the option:
d.) \( 2 e^{x^{2}+3} \)
should be checked.

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Would you like to ask any additional questions about this solution? Here are five questions to consider:

1. Can you explain another method to solve the integral \( \int 6 x^3 e^{x^2 + 3} \, dx \)?
2. How do you determine the choice of \( u \) and \( dv \) in integration by parts?
3. Could you demonstrate another example using integration by parts?
4. What are some common mistakes to avoid when applying integration by parts?
5. How can we verify the solution to an integral?

Tip: When choosing \( u \) and \( dv \) in integration by parts, selecting \( u \) such that \( du \) simplifies the integral can make the process easier.

Learn how to solve the integral of 6x^3 e^(x^2 + 3) using integration by parts with a detailed step-by-step solution. This problem involves calculus techniques including integration by parts and handling exponential functions. Suitable for undergraduate level students.

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