Let's consider the substitution \( u = x^2 \) instead of \( u = x^3 \) for the integral \(\int_0^1 x^2 e^{x^3} (1 + x^3) \, dx\).

First, let's rewrite the integral for clarity:

\[
\int_0^1 x^2 e^{x^3} (1 + x^3) \, dx
\]

Now, we'll use the substitution \( u = x^2 \). Then, \( du = 2x \, dx \), or equivalently, \( \frac{1}{2} du = x \, dx \).

Next, we need to adjust the limits of integration. When \( x = 0 \), \( u = 0^2 = 0 \). When \( x = 1 \), \( u = 1^2 = 1 \). So the new limits for \( u \) will be from 0 to 1.

Substituting these into the integral, we need to express \( x^2 \) and \( x^3 \) in terms of \( u \). Notice that \( x^2 = u \) and \( x^3 = u^{3/2} \).

The differential \( dx \) can be found from \( \frac{1}{2} du = x \, dx \), so \( dx = \frac{1}{2x} du = \frac{1}{2\sqrt{u}} du \).

Now, substitute everything into the integral:

\[
\int_0^1 x^2 e^{x^3} (1 + x^3) \, dx = \int_0^1 u e^{u^{3/2}} \left( 1 + u^{3/2} \right) \frac{1}{2\sqrt{u}} \, du
\]

Simplify the integrand:

\[
= \frac{1}{2} \int_0^1 u e^{u^{3/2}} \left( 1 + u^{3/2} \right) \frac{1}{\sqrt{u}} \, du
\]

\[
= \frac{1}{2} \int_0^1 e^{u^{3/2}} \left( \sqrt{u} + u^2 \right) \, du
\]

Split the integral into two parts:

\[
= \frac{1}{2} \left( \int_0^1 e^{u^{3/2}} \sqrt{u} \, du + \int_0^1 e^{u^{3/2}} u^2 \, du \right)
\]

Let’s denote these integrals as \( I_1 \) and \( I_2 \):

\[
I_1 = \int_0^1 e^{u^{3/2}} \sqrt{u} \, du
\]

\[
I_2 = \int_0^1 e^{u^{3/2}} u^2 \, du
\]

Now, consider the first integral \( I_1 \):

\[
I_1 = \int_0^1 e^{u^{3/2}} u^{1/2} \, du
\]

We use another substitution for this, \( v = u^{3/2} \). Then, \( dv = \frac{3}{2} u^{1/2} \, du \), or \( du = \frac{2}{3} u^{-1/2} \, dv \).

When \( u = 0 \), \( v = 0 \). When \( u = 1 \), \( v = 1^{3/2} = 1 \).

\[
I_1 = \int_0^1 e^v \frac{2}{3} u^{-1/2} u^{1/2} \, dv = \frac{2}{3} \int_0^1 e^v \, dv
\]

\[
= \frac{2}{3} \left( e^v \bigg|_0^1 \right) = \frac{2}{3} (e - 1)
\]

For the second integral \( I_2 \):

\[
I_2 = \int_0^1 e^{u^{3/2}} u^2 \, du
\]

Using the same substitution \( v = u^{3/2} \), \( du = \frac{2}{3} u^{-1/2} \, dv \):

\[
I_2 = \int_0^1 e^v \frac{2}{3} u^{4/3} u^{-1/2} \, dv = \frac{2}{3} \int_0^1 e^v u^{5/2-1/2} \, dv = \frac{2}{3} \int_0^1 e^v u^2 \, dv
\]

Since \( u = v^{2/3} \):

\[
= \frac{2}{3} \int_0^1 e^v v^{4/3} \, dv
\]

This integral does not simplify easily, so we stick with the original substitution \( u = x^3 \) for a straightforward solution.

Thus, the exact value of the integral remains:

\[
\boxed{\frac{1}{3} e}
\]

Would you like further details or have any questions?

Here are 5 more questions you might consider:
1. How do you perform integration by parts?
2. Can you explain substitution in integrals in more detail?
3. How do you solve definite integrals with variable limits?
4. What are some common techniques for evaluating complex integrals?
5. Can you solve \(\int x \sin(x^2) \, dx\)?

**Tip:** When performing substitutions, carefully track how variables and differentials change to ensure accurate results.

Explore how to solve the integral \(\int_0^1 x^2 e^{x^3} (1 + x^3) \, dx\) using the substitution \( u = x^2 \). This detailed guide covers substitution methods, integral evaluation, and advanced calculus techniques. Ideal for advanced high school and introductory college students.

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