(i) Let \(u = 1 + x^2\). Then \(du = 2x \, dx\), so \(dx = \frac{du}{2x}\).

Substitute \(x^2 = u - 1\) into the integral:

\(I = \int_0^1 \frac{x^5}{(1+x^2)^3} \, dx = \int_1^2 \frac{(u-1)^{5/2}}{u^3} \cdot \frac{du}{2\sqrt{u-1}}\).

Simplifying, we get:

\(I = \int_1^2 \frac{(u-1)^2}{2u^3} \, du\).

(ii) To find the exact value of \(I\), integrate:

\(\int \frac{(u-1)^2}{2u^3} \, du = \int \left( \frac{1}{2u} - \frac{1}{u^2} + \frac{1}{4u^2} \right) \, du\).

This gives:

\(\frac{1}{2} \ln u + \frac{1}{u} - \frac{1}{4u^2}\).

Evaluate from 1 to 2:

\(\left[ \frac{1}{2} \ln u + \frac{1}{u} - \frac{1}{4u^2} \right]_1^2 = \left( \frac{1}{2} \ln 2 + \frac{1}{2} - \frac{1}{16} \right) - \left( 0 + 1 - \frac{1}{4} \right)\).

Simplifying, we find:

\(\frac{1}{2} \ln 2 - \frac{5}{16}\).