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The integral provided is:

\[
I = \int_0^1 \frac{x^2}{\sqrt{1 + x^2}} \, dx
\]

To solve this integral, we can use the substitution \( u = 1 + x^2 \), so \( du = 2x \, dx \).

### Step-by-Step Solution:

1. **Substitution**:
   Let \( u = 1 + x^2 \), then \( du = 2x \, dx \).
   - When \( x = 0 \), \( u = 1 \).
   - When \( x = 1 \), \( u = 2 \).

   The integral becomes:
   \[
   I = \int_{u=1}^{u=2} \frac{x}{\sqrt{u}} \, du
   \]

   Since \( x = \sqrt{u - 1} \), substitute this:
   \[
   I = \int_{1}^{2} \frac{\sqrt{u - 1}}{\sqrt{u}} \cdot \frac{du}{2}
   \]

   Simplify:
   \[
   I = \frac{1}{2} \int_{1}^{2} \frac{\sqrt{u - 1}}{\sqrt{u}} \, du
   \]

2. **Simplify the Fraction**:
   Combine the radicals:
   \[
   I = \frac{1}{2} \int_{1}^{2} \sqrt{\frac{u - 1}{u}} \, du
   \]

   Rewrite the fraction inside the square root:
   \[
   \frac{u - 1}{u} = 1 - \frac{1}{u}
   \]

   So:
   \[
   I = \frac{1}{2} \int_{1}^{2} \sqrt{1 - \frac{1}{u}} \, du
   \]

3. **Approach for Integration**:
   At this stage, the integral involves a square root, which does not simplify further without numerical techniques or a series expansion. You can either proceed numerically or look for an approximate form.

Would you like me to calculate this integral numerically or explore further analytical techniques?

Evaluate the integral \( \int_0^1 \frac{x^2}{\sqrt{1 + x^2}} \, dx \).

Explore a detailed step-by-step solution for evaluating the integral \( \int_0^1 \frac{x^2}{\sqrt{1 + x^2}} \, dx \). This problem involves substitution techniques, simplification of radicals, and integral calculus concepts. Suitable for undergraduate calculus students.

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