By using the substitution $u = 1 + t^3$, find the exact value of $\int_{0}^{2} \frac{t^5}{\left(1 + t^3\right)^3} \text{ d}t$ without using a calculator.

1. Find the derivative of the substitution to relate $du$ and $dt$.

$u = 1 + t^3$

$\frac{\text{d}u}{\text{d}t} = 3t^2$

$t^2 \text{ d}t = \frac{1}{3} \text{ d}u$

2. Determine the new limits of integration in terms of $u$.

When $t = 0$, $u = 1 + 0^3 = 1$.

When $t = 2$, $u = 1 + 2^3 = 9$.

3. Rewrite the integrand entirely in terms of $u$.

Rewrite $t^5$ as $t^3 \cdot t^2$:

$t^3 = u - 1$

$\int_{0}^{2} \frac{t^5}{\left(1 + t^3\right)^3} \text{ d}t = \int_{0}^{2} \frac{t^3}{\left(1 + t^3\right)^3} \cdot t^2 \text{ d}t$

$= \int_{1}^{9} \frac{u - 1}{u^3} \left(\frac{1}{3}\right) \text{ d}u$

$= \frac{1}{3} \int_{1}^{9} \left( \frac{u}{u^3} - \frac{1}{u^3} \right) \text{ d}u$

$= \frac{1}{3} \int_{1}^{9} \left( u^{-2} - u^{-3} \right) \text{ d}u$

4. Integrate the function with respect to $u$.

$= \frac{1}{3} \left[ \frac{u^{-1}}{-1} - \frac{u^{-2}}{-2} \right]_{1}^{9}$

$= \frac{1}{3} \left[ -\frac{1}{u} + \frac{1}{2u^2} \right]_{1}^{9}$

5. Evaluate the definite integral using the upper and lower limits.

$= \frac{1}{3} \left[ \left( -\frac{1}{9} + \frac{1}{2(9^2)} \right) - \left( -\frac{1}{1} + \frac{1}{2(1^2)} \right) \right]$

$= \frac{1}{3} \left[ \left( -\frac{18}{162} + \frac{1}{162} \right) - \left( -1 + \frac{1}{2} \right) \right]$

$= \frac{1}{3} \left[ -\frac{17}{162} - \left( -\frac{1}{2} \right) \right]$

$= \frac{1}{3} \left[ -\frac{17}{162} + \frac{81}{162} \right]$

$= \frac{1}{3} \left[ \frac{64}{162} \right] = \frac{1}{3} \left[ \frac{32}{81} \right] = \frac{32}{243}$