[Cambridge/9233/2004] Use the substitution $t = \sqrt{(x + 1)}$ to find the exact value of $\int_{0}^{3} x \sqrt{(x + 1)} \, dx$.

1. Express $x$ and $dx$ in terms of $t$ and $dt$.

$t = \sqrt{x + 1}$
$t^2 = x + 1 \implies x = t^2 - 1$
Differentiating with respect to $t$:
$dx = 2t \, dt$

2. Determine the new limits of integration and set up the new integral.

When $x = 0$, $t = \sqrt{0 + 1} = 1$.
When $x = 3$, $t = \sqrt{3 + 1} = 2$.

Substitute these into the integral:
$\int_{0}^{3} x \sqrt{x + 1} \, dx = \int_{1}^{2} (t^2 - 1)(t)(2t) \, dt$

3. Expand the integrand and integrate with respect to $t$.

$\int_{1}^{2} 2t^2(t^2 - 1) \, dt = \int_{1}^{2} (2t^4 - 2t^2) \, dt$

Integrating term by term:
$= \left[ \frac{2t^5}{5} - \frac{2t^3}{3} \right]_{1}^{2}$

4. Evaluate the definite integral to find the exact value.

$= \left( \frac{2(2)^5}{5} - \frac{2(2)^3}{3} \right) - \left( \frac{2(1)^5}{5} - \frac{2(1)^3}{3} \right)$
$= \left( \frac{64}{5} - \frac{16}{3} \right) - \left( \frac{2}{5} - \frac{2}{3} \right)$
$= \left( \frac{192 - 80}{15} \right) - \left( \frac{6 - 10}{15} \right)$
$= \frac{112}{15} - \left( -\frac{4}{15} \right)$
$= \frac{116}{15}$