To solve the integral \( \int x^3 e^{x^2} \, dx \), we can use the method of substitution.

### Step 1: Substitution
Let:
\[
u = x^2
\]
Then, differentiate both sides with respect to \( x \):
\[
du = 2x \, dx \quad \text{or} \quad \frac{du}{2} = x \, dx
\]

### Step 2: Rewrite the integral
Now, rewrite the original integral in terms of \( u \). Since \( x^3 = x \cdot x^2 = x \cdot u \), we substitute \( u = x^2 \) and \( x \, dx = \frac{du}{2} \):
\[
\int x^3 e^{x^2} \, dx = \int x \cdot u \cdot e^u \cdot \frac{du}{2}
\]
\[
= \frac{1}{2} \int u e^u \, du
\]

### Step 3: Use integration by parts
Now, apply integration by parts. Let:
\[
v = u \quad \text{and} \quad dw = e^u \, du
\]
Differentiate and integrate:
\[
dv = du \quad \text{and} \quad w = e^u
\]
Now, apply the formula for integration by parts:
\[
\int v \, dw = v \cdot w - \int w \, dv
\]
Substituting the values:
\[
\frac{1}{2} \int u e^u \, du = \frac{1}{2} \left( u e^u - \int e^u \, du \right)
\]
\[
= \frac{1}{2} \left( u e^u - e^u \right) = \frac{1}{2} e^u (u - 1)
\]

### Step 4: Substitute back in terms of \( x \)
Recall that \( u = x^2 \), so:
\[
\frac{1}{2} e^u (u - 1) = \frac{1}{2} e^{x^2} (x^2 - 1)
\]

### Final Answer:
Thus, the solution to the integral is:
\[
\int x^3 e^{x^2} \, dx = \frac{1}{2} e^{x^2} (x^2 - 1) + C
\]
where \( C \) is the constant of integration.

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. How would you solve \( \int x e^{x^2} \, dx \)?
2. Can you apply integration by parts to \( \int x^2 e^x \, dx \)?
3. What is the derivative of \( \frac{1}{2} e^{x^2} (x^2 - 1) \)?
4. How do you handle integrals involving \( e^{x^2} \) without a polynomial?
5. What is the solution to \( \int e^{x^2} \, dx \)?

**Tip:** When using substitution, always check if the new variables simplify the integral and avoid unnecessary complexity.

Learn how to solve the integral ∫x^3 e^{x^2} dx using substitution and integration by parts. This step-by-step guide covers key concepts like exponential functions, substitution method, and integration by parts. Suitable for students in Calculus II.

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