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The problem asks to evaluate the integral:

\[
\int x^9 e^{2x^5} \, dx
\]

### Approach: 
We can solve this integral using substitution. Let:

\[
u = 2x^5
\]

Then, the derivative of \( u \) with respect to \( x \) is:

\[
du = 10x^4 \, dx \quad \Rightarrow \quad \frac{du}{10} = x^4 \, dx
\]

Now, rewrite the integral:

\[
\int x^9 e^{2x^5} \, dx = \int x^5 \cdot x^4 e^{2x^5} \, dx = \int x^5 e^{u} \cdot \frac{du}{10}
\]

Since \( x^5 = \frac{u}{2} \), we can substitute and simplify:

\[
\int x^9 e^{2x^5} \, dx = \frac{1}{10} \int u e^u \, du
\]

### Integration by Parts:
We use integration by parts for \( \int u e^u \, du \). Let:

\[
v = u, \quad dw = e^u \, du
\]
\[
dv = du, \quad w = e^u
\]

Applying the integration by parts formula:

\[
\int v \, dw = vw - \int w \, dv
\]
\[
\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u + C
\]

Substitute back for \( u = 2x^5 \):

\[
\frac{1}{10} \int u e^u \, du = \frac{1}{10} \left( 2x^5 e^{2x^5} - e^{2x^5} \right) + C
\]

Factoring out \( e^{2x^5} \):

\[
= \frac{1}{10} e^{2x^5} \left( 2x^5 - 1 \right) + C
\]

### Final Answer:
The correct choice is **A**:

\[
\frac{1}{10} e^{2x^5} \left( 2x^5 - 1 \right) + C
\]

Would you like further explanation on integration by parts or substitution methods?

Here are five related questions to consider:
1. How would the result change if the exponent in \( e^{2x^5} \) was different?
2. What happens if we integrate \( x^9 e^{kx^5} \) for a general constant \( k \)?
3. Can you solve a similar problem where \( e^{x^n} \) appears in the integral?
4. How does integration by parts help solve non-trivial integrals like this one?
5. How would you handle an integral with a higher power of \( x \)?

**Tip:** Always check if substitution simplifies the integrand before applying more complex techniques.

Evaluate the integral \(\int x^9 e^{2x^5} dx\).

Learn how to solve the integral \(\int x^9 e^{2x^5} dx\) using substitution and integration by parts. This problem covers key calculus concepts, including the substitution method and integration by parts. Suitable for students in Grades 11-12 and early university levels.

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