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?